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GCSE Maths · EdexcelChapter 11 · Sequences
Chapter 11 · Higher tier

GCSE Maths Sequences and nth Term (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

  • Continue a sequence using a term-to-term rule, and describe the rule in words
  • Generate terms from a position-to-term rule (an nth-term formula)
  • Continue and describe patterns made from matchsticks, tiles and dots
  • Find the nth term of an arithmetic (linear) sequence using the common difference
  • Decide whether a number is a term of a sequence, and find the first term above or below a value
  • Recognise square, cube, triangular, Fibonacci-type and geometric sequences
  • Higher: find the nth term of a quadratic sequence using second differences
  • Higher: find the nth term of a geometric sequence, including sequences with a surd common ratio

You should already know

  • Adding, subtracting, multiplying and dividing whole numbers, decimals and negative numbers (Chapter 1)
  • Square numbers, cube numbers and index notation (Chapter 2)
  • The order of operations, BIDMAS (Chapter 1)
  • Collecting terms and using letters in algebra (Chapter 7)

Where this leads

  • Straight-line graphs: the nth term is like y = mx + c (Chapter 12)
  • Quadratic graphs, which have the same shape as quadratic sequences (Chapter 13)
  • Solving quadratic equations (Chapter 14)
  • Compound interest and growth are geometric sequences (Chapters 5 and 25)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: A23, A24, A25.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

Exercise P  Do you remember?

Non-calculator
1
FWork out 4 × 6 − 3.
2
FWrite down the first five multiples of 7.
3
FWork out (a) 6² and (b) 2³.
(a)6²
(b)2³
4
FWork out 20 − 3 × 4.
5
FWork out (a) −7 + 12 and (b) 3 − 8.
(a)−7 + 12
(b)3 − 8
11.1

Sequences and term-to-term rules

Edexcel 1MA1: A23
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Sequences and term-to-term rulesComing soon

A sequence is a list of numbers, in a fixed order, that follow a rule. You meet sequences all the time. A bus leaves at 9:00, 9:20, 9:40, … A phone plan costs £10 in month 1, £14 in month 2, £18 in month 3, … Each number in the list is called a term. We say the first term, the second term, the third term, and so on. A sequence that gets bigger is increasing (ascending) and one that gets smaller is decreasing (descending).

The simplest way to describe a sequence is the term-to-term rule. This tells you what to do to one term to get the next term. In the sequence 3, 7, 11, 15, … you add 4 each time, so the term-to-term rule is “add 4”. The arrows below show it.

371115+4+4+4…
The term-to-term rule is “add 4”. Each arrow shows the rule being used once.

To find the rule, look at how each term changes into the next. Always try these checks in this order:

  1. Differences. Subtract each term from the one after it. If the answer is always the same, the rule is “add that number” (or “subtract” if the answers are negative).
  2. Ratios. If the differences are not the same, divide each term by the one before it. If the answer is always the same, the rule is “multiply by that number” (or “divide by” when the sequence shrinks).
  3. Changing differences. Sometimes the amount you add changes in a regular way, for example +1, +2, +3, …
24816×2×2×2…
Here the rule is “multiply by 2”: each term is double the one before.

Some rules have two steps, such as “multiply by 2 and then add 1”. Starting at 3 you get 3, then 3 × 2 + 1 = 7, then 7 × 2 + 1 = 15, and so on. When a rule is given to you, the question will also tell you the first term, because a rule alone cannot tell you where to start.

Finally, to continue a sequence, always apply the rule to the last term you have. Do not go back to the start each time.

Key ideas

  • A sequence is an ordered list of numbers; each number is a term.
  • The term-to-term rule says what to do to one term to get the next.
  • Check the differences first (add or subtract), then the ratios (multiply or divide).
  • Sometimes the difference itself changes by the same amount each time (+1, +2, +3, …).
  • To continue a sequence, apply the rule to the last term.
Worked example 11.1.1
Write down the next two terms of the sequence 5, 9, 13, 17, … and describe the term-to-term rule.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 2 terms.

  1. Find the difference between each pair of neighbouring terms. Subtract the earlier term from the later term. A sequence that goes up has positive differences; one that goes down has negative differences.
    9 − 5 = 4
    13 − 9 = 4
    17 − 13 = 4
  2. Look at the differences. They are all the same, so the same number is added each time (a negative difference means we subtract).
    Every difference is 4, so the rule is add 4.
  3. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    17 + 4 = 21
    21 + 4 = 25
  4. Write down the new terms.
    21, 25

Check. Jump straight there: 5 + 5 × 4 = 25, and the last term we found is 25. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Worked example 11.1.2
The sequence 50, 44, 38, 32, … is going down. Write down the next two terms.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 2 terms.

  1. Find the difference between each pair of neighbouring terms. Subtract the earlier term from the later term. A sequence that goes up has positive differences; one that goes down has negative differences.
    44 − 50 = −6
    38 − 44 = −6
    32 − 38 = −6
  2. Look at the differences. They are all the same, so the same number is added each time (a negative difference means we subtract).
    Every difference is −6, so the rule is subtract 6.
  3. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    32 − 6 = 26
    26 − 6 = 20
  4. Write down the new terms.
    26, 20

Check. Jump straight there: 50 − 5 × 6 = 20, and the last term we found is 20. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Worked example 11.1.3
Write down the next two terms of 3, 6, 12, 24, … and describe the rule.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 2 terms.

  1. Try the differences first. We always look for "add or subtract the same number" before anything else.
    6 − 3 = 3
    12 − 6 = 6
    24 − 12 = 12
    The differences are not all the same, so it is not an add/subtract rule.
  2. Now divide each term by the term before it. If we are multiplying by the same number each time, dividing a term by the one before gives that number.
    6 ÷ 3 = 2
    12 ÷ 6 = 2
    24 ÷ 12 = 2
  3. Look at the answers. They are all the same, so we multiply by that number each time.
    The rule is multiply by 2.
  4. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    24 × 2 = 48
    48 × 2 = 96
  5. Write down the new terms.
    48, 96

Check. Look back along the whole list: 12, 24, 48, 96. Every term is the one before multiplied by 2. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Worked example 11.1.4
The first term of a sequence is 2. The rule is “multiply by 2 and then add 1”. Write down the first five terms.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. A term-to-term rule tells you what to do to one term to get the next. Start with the first term and apply the rule again and again.

  1. Write down the first term. The first term is given, so it is the start of the list.
    Term 1 = 2
  2. Apply the rule "multiply by 2 and then add 1" to find each new term. Each term comes from the term before it. Use the answer you have just found as the next starting point.
    Term 2: 2 × 2 = 4, then 4 + 1 = 5
    Term 3: 5 × 2 = 10, then 10 + 1 = 11
    Term 4: 11 × 2 = 22, then 22 + 1 = 23
    Term 5: 23 × 2 = 46, then 46 + 1 = 47
  3. Write the terms as a list.
    2, 5, 11, 23, 47

Check. Test the rule on the last pair: 23 × 2, then + 1 gives 47. ✓

Watch out. Do the operations in the order the rule gives them. "Multiply by 2 then add 1" is not the same as "add 1 then multiply by 2".

Worked example 11.1.5
Write down the next two terms of 1, 2, 4, 7, 11, …

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 2 terms.

  1. Try the differences first. We always look for "add or subtract the same number" before anything else.
    2 − 1 = 1
    4 − 2 = 2
    7 − 4 = 3
    11 − 7 = 4
    The differences are not the same.
  2. Look at how the differences change. Sometimes the amount added grows by the same amount each time.
    1, 2, 3, 4
    The differences go up by 1 each time.
  3. Write the rule.
    The differences increase by 1 each time: add 1, then add 2, then add 3, and so on.
  4. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    11 + 5 = 16   (the next difference is 5)
    16 + 6 = 22   (the next difference is 6)
  5. Write down the new terms.
    16, 22

Check. Look at the differences of the whole list: 1, 2, 3, 4, 5, 6. They keep increasing by 1. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Watch out

The rule “add 3” does not tell you where the sequence starts: 2, 5, 8, … and 10, 13, 16, … both use it. And be careful not to describe 3, 6, 12, 24 as “add 3”: the first step adds 3 but the second adds 6. A rule must work for every pair of neighbouring terms.

Exercise 11A  Term-to-term rules

Non-calculator
1
FWrite down the next two terms of each sequence.
(a)4, 7, 10, 13, …
(b)30, 26, 22, 18, …
(c)5, 10, 20, 40, …
(d)64, 32, 16, 8, …
(e)1, 2, 4, 7, 11, …
(f)0.5, 1.2, 1.9, 2.6, …
2
FDescribe the term-to-term rule of each sequence in words.
(a)6, 11, 16, 21, …
(b)100, 93, 86, 79, …
(c)3, 6, 12, 24, …
(d)162, 54, 18, 6, …
(e)2.5, 3, 3.5, 4, …
(f)−4, −1, 2, 5, …
3
FThe first term and the term-to-term rule of some sequences are given. Write down the first five terms of each.
(a)First term 4; rule: add 6
(b)First term 100; rule: subtract 7
(c)First term 3; rule: multiply by 2
(d)First term 80; rule: divide by 2
(e)First term 2; rule: multiply by 3 and then subtract 1
(f)First term 1; rule: multiply by 2 and then add 3
4
PSSam has £8 in his money box in week 1. Each week he adds £5. How much does he have in week 6?
5
FFind the 10th term of the sequence 3, 8, 13, 18, …
6
RAli says, “The rule for 3, 6, 12, 24, … is add 3.” Explain why Ali is wrong.
7
FFind the missing numbers in each sequence.
(a)5 ,  ?  , 17 , 23 ,  ? 
(b)100 , 88 ,  ?  , 64 ,  ? 
(c) ?  , 14 , 21 ,  ?  , 35
(d)-12 ,  ?  , -4 , 0 ,  ? 
8
FWrite down the next two terms of each sequence.
(a)14, 12, 34, 1, …
(b)12, 1, 32, 2, …
(c)8, 5, 2, −1, …
(d)−20, −15, −10, −5, …
9
RFreya says the next term in the sequence 1, 2, 4, 7, 11, … is 15 because “the last difference was 4, so add 4 again”. Is she right? Explain.
10
PSA tank holds 640 litres at the start. Each day half of the water is used. Term 1 is the amount at the start, term 2 is the amount after 1 day, and so on. How much water is left after 4 days?
11
PSThe first term of a sequence is 5. The rule is “multiply by 2 and then subtract 3”. Work out the 7th term.
12
RThe first term of a sequence is 2 and the rule is “multiply by 2 and then subtract 1”. Work out the first term that is bigger than 1000.
11.2

Patterns and diagrams

Edexcel 1MA1: A23, A24
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Patterns and diagramsComing soon

Many sequences come from patterns of shapes, made with matchsticks, tiles or dots. Each pattern in the list uses a few more than the one before. To turn a pattern into a sequence of numbers there are four steps:

  1. Count how many matchsticks (or tiles, or dots) are in each pattern.
  2. Write the counts as a list. That list is a sequence.
  3. Find the rule that takes you from one count to the next.
  4. Use the rule to carry on and predict later patterns.

Here are the first four patterns made from matchsticks. Count the sticks: 4, 7, 10, 13. The rule is “add 3”.

Pattern 1Pattern 2Pattern 3Pattern 4
Rows of squares made from matchsticks. The counts are 4, 7, 10, 13.

It is very important to see why the rule works by looking at the picture. Each new square is stuck on the end of the row. It shares one side with the square before it, so it needs only 3 new matchsticks, not 4. That is why the rule is “add 3”. If you can explain the rule from the picture, you can trust it.

Sometimes the number you add changes. Look at these patterns of dots. The first has 1, 3, 6, 10 dots; the differences are 2, 3, 4, so each pattern adds one more dot than the last pattern added. These are called the triangular numbers. The second pattern of dots has 1, 4, 9, 16 dots. The differences are 3, 5, 7. These are the square numbers, because the dots make a square.

Pattern 1Pattern 2Pattern 3Pattern 4
Triangular patterns of dots: 1, 3, 6, 10. The number added goes up by 1 each time.
Pattern 1Pattern 2Pattern 3Pattern 4
Square patterns of dots: 1, 4, 9, 16. The number added goes up by 2 each time.

You can carry on a rule for a pattern 5th, 6th, 10th and so on. When you want a very large pattern, such as pattern 100, continuing one step at a time takes far too long. In sections 11.4 and 11.5 you will learn to write a formula (the nth term) that jumps straight to any pattern.

Key ideas

  • To turn a pattern into a sequence: count, write the list, find the rule, carry on.
  • Explain the rule with the picture: new shapes often share a side, so they add fewer than a whole shape.
  • If the numbers added go up by the same amount each time, the sequence is not “add the same number”: look at the differences between differences.
  • Triangular numbers: 1, 3, 6, 10, …. Square numbers: 1, 4, 9, 16, …
Worked example 11.2.1
Here are the first three patterns made from matchsticks.
Pattern 1Pattern 2Pattern 3
How many matchsticks are needed for pattern 6?

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Count the items in each pattern to make a sequence, find the term-to-term rule, then continue the sequence.

  1. Count the matchsticks in each pattern shown. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
    Pattern123
    Number of matchsticks4710
  2. Find the difference between each pair of neighbouring terms. Subtract the earlier term from the later term. A sequence that goes up has positive differences; one that goes down has negative differences.
    7 − 4 = 3
    10 − 7 = 3
  3. Look at the differences. They are all the same, so the same number is added each time (a negative difference means we subtract).
    Every difference is 3, so the rule is add 3.
  4. Use the rule to carry on until you reach the pattern you want. Each pattern follows from the one before, so keep applying the rule.
    Pattern 4: 10 + 3 = 13
    Pattern 5: 13 + 3 = 16
    Pattern 6: 16 + 3 = 19
  5. Write the answer.
    Pattern 6 has 19 matchsticks.

Check. Look at the drawing: The first square needs 4 matchsticks. Each new square is added on the end and shares one side with the square before it, so it needs only 3 new matchsticks. After n squares there are 3 matchsticks for each square, plus 1 extra for the very first side: 3n + 1.

Watch out. Pattern 10 does not have 10 times as many matchsticks as pattern 1. The pattern grows by adding, so check the rule against the counts you already have.

Worked example 11.2.2
Here are the first three patterns of triangles.
Pattern 1Pattern 2Pattern 3
How many matchsticks are needed for pattern 8?

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Count the items in each pattern to make a sequence, find the term-to-term rule, then continue the sequence.

  1. Count the matchsticks in each pattern shown. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
    Pattern123
    Number of matchsticks357
  2. Find the difference between each pair of neighbouring terms. Subtract the earlier term from the later term. A sequence that goes up has positive differences; one that goes down has negative differences.
    5 − 3 = 2
    7 − 5 = 2
  3. Look at the differences. They are all the same, so the same number is added each time (a negative difference means we subtract).
    Every difference is 2, so the rule is add 2.
  4. Use the rule to carry on until you reach the pattern you want. Each pattern follows from the one before, so keep applying the rule.
    Pattern 4: 7 + 2 = 9
    Pattern 5: 9 + 2 = 11
    Pattern 6: 11 + 2 = 13
    Pattern 7: 13 + 2 = 15
    Pattern 8: 15 + 2 = 17
  5. Write the answer.
    Pattern 8 has 17 matchsticks.

Check. Look at the drawing: The first triangle needs 3 matchsticks. Each new triangle shares a side with the last one, so it needs only 2 new matchsticks. After n triangles there are 2 matchsticks for each triangle, plus 1 extra for the first one: 2n + 1.

Watch out. Pattern 10 does not have 10 times as many matchsticks as pattern 1. The pattern grows by adding, so check the rule against the counts you already have.

Worked example 11.2.3
Here are the first three tile patterns.
Pattern 1Pattern 2Pattern 3
How many tiles are in pattern 6?

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Count the items in each pattern to make a sequence, find the term-to-term rule, then continue the sequence.

  1. Count the tiles in each pattern shown. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
    Pattern123
    Number of tiles5913
  2. Find the difference between each pair of neighbouring terms. Subtract the earlier term from the later term. A sequence that goes up has positive differences; one that goes down has negative differences.
    9 − 5 = 4
    13 − 9 = 4
  3. Look at the differences. They are all the same, so the same number is added each time (a negative difference means we subtract).
    Every difference is 4, so the rule is add 4.
  4. Use the rule to carry on until you reach the pattern you want. Each pattern follows from the one before, so keep applying the rule.
    Pattern 4: 13 + 4 = 17
    Pattern 5: 17 + 4 = 21
    Pattern 6: 21 + 4 = 25
  5. Write the answer.
    Pattern 6 has 25 tiles.

Check. Look at the drawing: There is always 1 tile in the middle. Then there is an arm going up, down, left and right, and each arm has n tiles. So the number of tiles is 4 arms of n tiles plus the 1 in the middle: 4n + 1.

Watch out. Pattern 10 does not have 10 times as many tiles as pattern 1. The pattern grows by adding, so check the rule against the counts you already have.

Worked example 11.2.4
Here are the first four patterns of dots.
Pattern 1Pattern 2Pattern 3Pattern 4
How many dots are in pattern 7?

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Count the items in each pattern to make a sequence, find the term-to-term rule, then continue the sequence.

  1. Count the dots in each pattern shown. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
    Pattern1234
    Number of dots13610
  2. Try the differences first. We always look for "add or subtract the same number" before anything else.
    3 − 1 = 2
    6 − 3 = 3
    10 − 6 = 4
    The differences are not the same.
  3. Look at how the differences change. Sometimes the amount added grows by the same amount each time.
    2, 3, 4
    The differences go up by 1 each time.
  4. Write the rule.
    The differences increase by 1 each time: add 2, then add 3, then add 4, and so on.
  5. Use the rule to carry on until you reach the pattern you want. Each pattern follows from the one before, so keep applying the rule.
    Pattern 5: 10 + 5 = 15
    Pattern 6: 15 + 6 = 21
    Pattern 7: 21 + 7 = 28
  6. Write the answer.
    Pattern 7 has 28 dots.

Check. Look at the drawing: Pattern n has rows of 1, 2, 3, … up to n dots, so the number of dots is 1 + 2 + 3 + … + n. Two copies make a rectangle with n rows and n + 1 columns, so one copy has n(n + 1) ÷ 2 dots.

Watch out. Pattern 10 does not have 10 times as many dots as pattern 1. The pattern grows by adding, so check the rule against the counts you already have.

Worked example 11.2.5
Pattern 1, 2, 3 of the square matchstick pattern use 4, 7, 10 matchsticks. Is there a pattern that uses exactly 31 matchsticks?

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Make a sequence from the counts, then keep adding the common difference to see if the number is reached.

  1. Count the matchsticks in each pattern shown. Turn the pictures into a sequence of numbers.
    Pattern123
    Number of matchsticks4710
  2. Find the term-to-term rule. Subtract each count from the next.
    7 − 4 = 3
    10 − 7 = 3
    Rule: add 3
  3. Carry on adding until you reach or pass the number. The counts keep going up, so stop as soon as the count is 31 or bigger.
    P1: 4, P2: 7, P3: 10, P4: 13, P5: 16, P6: 19, P7: 22, P8: 25, P9: 28, P10: 31
  4. Decide. 31 appears in the list.
    Yes: pattern 10 has exactly 31 matchsticks.

Check. Pattern 10 follows from 4 by adding 3 a total of 9 times: 4 + 9 × 3 = 31. ✓

Watch out. Stop when you reach or pass the number. If the counts jump over it, there is no such pattern.

Watch out

Do not just multiply. Pattern 3 of the squares uses 10 matchsticks, but pattern 6 does not use 20: it uses 19. The patterns grow by adding the same amount, so doubling the pattern number does not double the count.

Exercise 11B  Patterns and diagrams

Non-calculator
1
FHere are the first three patterns made from matchsticks.
Pattern 1Pattern 2Pattern 3
(a)How many matchsticks are needed for pattern 4?
(b)How many matchsticks are needed for pattern 5?
(c)How many matchsticks are needed for pattern 8?
2
FHere are the first three patterns made from matchsticks.
Pattern 1Pattern 2Pattern 3
(a)How many matchsticks are needed for pattern 4?
(b)How many matchsticks are needed for pattern 5?
(c)How many matchsticks are needed for pattern 9?
3
FHere are the first three tile patterns.
Pattern 1Pattern 2Pattern 3
(a)How many tiles are needed for pattern 4?
(b)How many tiles are needed for pattern 5?
(c)How many tiles are needed for pattern 7?
4
FHere are the first three patterns made from matchsticks.
Pattern 1Pattern 2Pattern 3
(a)How many matchsticks are needed for pattern 4?
(b)How many matchsticks are needed for pattern 6?
(c)How many matchsticks are needed for pattern 8?
5
PSA path is made of blue tiles in the middle with orange tiles around the border. Here are the first three patterns.
Pattern 1Pattern 2Pattern 3
(a)How many orange tiles are needed for pattern 4?
(b)How many orange tiles are needed for pattern 6?
(c)How many orange tiles are needed for pattern 10?
6
FHere are the first four patterns of dots.
Pattern 1Pattern 2Pattern 3Pattern 4
(a)How many dots are in pattern 5?
(b)How many dots are in pattern 6?
(c)How many dots are in pattern 8?
7
FHere are the first four patterns of dots.
Pattern 1Pattern 2Pattern 3Pattern 4
(a)How many dots are in pattern 5?
(b)How many dots are in pattern 6?
(c)How many dots are in pattern 8?
8
FHere are the first three patterns of squares made with matchsticks.
Pattern 1Pattern 2Pattern 3
(a)How many matchsticks are needed for pattern 4?
(b)How many matchsticks are needed for pattern 5?
(c)How many matchsticks are needed for pattern 7?
9
PSFor each pattern of squares or triangles, is there a pattern that uses exactly the number of matchsticks given?
(a)The squares pattern (4, 7, 10, …): exactly 31 matchsticks
(b)The squares pattern (4, 7, 10, …): exactly 40 matchsticks
(c)The squares pattern (4, 7, 10, …): exactly 50 matchsticks
(d)The triangles pattern (3, 5, 7, …): exactly 21 matchsticks
(e)The triangles pattern (3, 5, 7, …): exactly 30 matchsticks
10
PSTables at a school party are pushed together in a long row. One table seats 6 people, two tables seat 10 people, three tables seat 14 people, and so on. How many people can sit at a row of 8 tables?
11
RRob says, “Pattern 1 of the square matchstick pattern uses 4 matchsticks, so pattern 10 uses 10 × 4 = 40.” Is Rob right? Explain.
12
RA pattern uses 4, 9, 14, 19 matchsticks for patterns 1 to 4. Mia says there is a pattern that uses exactly 100 matchsticks. Is Mia right? Give a reason.
13
RIn the dot patterns that give the triangular numbers (1, 3, 6, 10, …), pattern 6 has 21 dots. How many more dots are needed to change pattern 6 into pattern 7? Explain how you know without counting all the dots.
14
RSam makes a row of 9 squares from matchsticks (as in the pattern above). Tara makes a row of 9 triangles from matchsticks. Who uses more matchsticks, and how many more?
11.3

Position-to-term rules

Edexcel 1MA1: A23
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Position-to-term rulesComing soon

A term-to-term rule is slow if you want a term that is far along. To find the 100th term you would need the 99 terms before it. A position-to-term rule (also called the nth term) is much quicker. It lets you find a term directly from its position.

The position of a term is its place in the list: 1st, 2nd, 3rd, and so on. We use the letter n for the position. The rule is a formula with n in it. For example, “nth term = 3n + 2” means “multiply the position by 3, then add 2”. The picture shows this as a machine.

n = 4× 312+ 214
The nth term 3n + 2 as a function machine. Position 4 goes in and the 4th term, 14, comes out.

To find a term, replace n by the position number and work it out. With 3n + 2:

Position n1234
3n + 23 × 1 + 2 = 53 × 2 + 2 = 83 × 3 + 2 = 113 × 4 + 2 = 14

So the sequence is 5, 8, 11, 14, … Use BIDMAS (Chapter 1) when you work it out: brackets and powers first, then multiplying and dividing, then adding and subtracting. The most common mistakes are with powers:

Terms can be negative. For 10 − 2n the terms are 8, 6, 4, 2, 0, −2, … (n = 6 gives 10 − 12 = −2). Some nth terms are special: n² gives the square numbers 1, 4, 9, 16, …; n³ gives the cube numbers 1, 8, 27, 64, …; 2n gives the powers of 2: 2, 4, 8, 16, … You will see these in Section 11.6.

Look back at 5, 8, 11, 14 from 3n + 2. The term-to-term rule is “add 3”, and the nth term has 3n in it. That is not an accident. The number in front of n is the amount the sequence goes up each time, which is the secret of Section 11.4.

Key ideas

  • The position n is the place of the term: the first term has n = 1.
  • To find a term, replace n by the position number and work it out using BIDMAS.
  • n² = n × n, and 2n² = 2 × (n × n).
  • A position-to-term rule jumps straight to any term, such as the 50th or 100th.
Worked example 11.3.1
The nth term of a sequence is 5n − 3. Write down the first four terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the formula by each position number, and work out the answer.

  1. Write the rule and say what it means. The letter n stands for the position of the term (1 for the first term, 2 for the second, and so on). To get a term, replace n by the position number.
    nth term = 5n − 3
  2. Replace n by 1 to find the 1st term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    1st term = 5 × 1 − 3 = 5 − 3 = 2
  3. Replace n by 2 to find the 2nd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    2nd term = 5 × 2 − 3 = 10 − 3 = 7
  4. Replace n by 3 to find the 3rd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    3rd term = 5 × 3 − 3 = 15 − 3 = 12
  5. Replace n by 4 to find the 4th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    4th term = 5 × 4 − 3 = 20 − 3 = 17
  6. Write the terms as a list.
    2, 7, 12, 17

Check. The terms should go up by 5 each time (the number in front of n): 5, 5, 5. ✓

Watch out. n² means n × n, not 2 × n. And the first term is when n = 1, not n = 0.

Worked example 11.3.2
The nth term of a sequence is n² + 4. Write down the first five terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the formula by each position number, and work out the answer.

  1. Write the rule and say what it means. The letter n stands for the position of the term (1 for the first term, 2 for the second, and so on). To get a term, replace n by the position number.
    nth term = n2 + 4
  2. Replace n by 1 to find the 1st term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    1st term = 12 + 4 = 1 + 4 = 5
  3. Replace n by 2 to find the 2nd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    2nd term = 22 + 4 = 4 + 4 = 8
  4. Replace n by 3 to find the 3rd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    3rd term = 32 + 4 = 9 + 4 = 13
  5. Replace n by 4 to find the 4th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    4th term = 42 + 4 = 16 + 4 = 20
  6. Replace n by 5 to find the 5th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    5th term = 52 + 4 = 25 + 4 = 29
  7. Write the terms as a list.
    5, 8, 13, 20, 29

Check. The first differences are 3, 5, 7, 9; the second differences are all 2, which is 2 × 1 as expected. ✓

Watch out. n² means n × n, not 2 × n. And the first term is when n = 1, not n = 0.

Worked example 11.3.3
The nth term of a sequence is 2n² − 1. Work out the 6th term.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the nth-term rule by the position you want.

  1. Write the rule. This formula gives the term for any position n.
    nth term = 2n2 − 1
  2. The position we want is 6, so replace n by 6. Every n in the formula becomes the same number.
    6th term = 2 × 62 − 1
  3. Work it out, doing powers first, then multiplying, then adding or subtracting. This is the BIDMAS order.
    6th term = 2 × 62 − 1 = 2 × 36 − 1 = 72 − 1 = 71
  4. Write the answer.
    The 6th term is 71.

Check. Check on a calculator: the 6th term is 71. ✓

Watch out. Substitute the position number for n. For 3n + 2 with n = 50 you work out 3 × 50 + 2 = 152, you do not multiply anything else by 50.

Worked example 11.3.4
The nth term of a sequence is 4n + 7. Work out the 100th term.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the nth-term rule by the position you want.

  1. Write the rule. This formula gives the term for any position n.
    nth term = 4n + 7
  2. The position we want is 100, so replace n by 100. Every n in the formula becomes the same number.
    100th term = 4 × 100 + 7
  3. Work it out, doing powers first, then multiplying, then adding or subtracting. This is the BIDMAS order.
    100th term = 4 × 100 + 7 = 400 + 7 = 407
  4. Write the answer.
    The 100th term is 407.

Check. The 100th term should be the first term plus 99 steps of 4: 11 + 99 × 4 = 407. ✓

Watch out. Substitute the position number for n. For 3n + 2 with n = 50 you work out 3 × 50 + 2 = 152, you do not multiply anything else by 50.

Worked example 11.3.5
The nth term of a sequence is 20 − 3n. Write down the first six terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the formula by each position number, and work out the answer.

  1. Write the rule and say what it means. The letter n stands for the position of the term (1 for the first term, 2 for the second, and so on). To get a term, replace n by the position number.
    nth term = 20 − 3n
  2. Replace n by 1 to find the 1st term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    1st term = 20 − 3 × 1 = 20 − 3 = 17
  3. Replace n by 2 to find the 2nd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    2nd term = 20 − 3 × 2 = 20 − 6 = 14
  4. Replace n by 3 to find the 3rd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    3rd term = 20 − 3 × 3 = 20 − 9 = 11
  5. Replace n by 4 to find the 4th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    4th term = 20 − 3 × 4 = 20 − 12 = 8
  6. Replace n by 5 to find the 5th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    5th term = 20 − 3 × 5 = 20 − 15 = 5
  7. Replace n by 6 to find the 6th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    6th term = 20 − 3 × 6 = 20 − 18 = 2
  8. Write the terms as a list.
    17, 14, 11, 8, 5, 2

Check. The terms should go up by −3 each time (the number in front of n): −3, −3, −3, −3, −3. ✓

Watch out. n² means n × n, not 2 × n. And the first term is when n = 1, not n = 0.

Worked example 11.3.6
The nth term of a sequence is n² − 3n + 5. Write down the first five terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the formula by each position number, and work out the answer.

  1. Write the rule and say what it means. The letter n stands for the position of the term (1 for the first term, 2 for the second, and so on). To get a term, replace n by the position number.
    nth term = n2 − 3n + 5
  2. Replace n by 1 to find the 1st term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    1st term = 12 − 3 × 1 + 5 = 1 − 3 × 1 + 5 = 1 − 3 + 5 = 3
  3. Replace n by 2 to find the 2nd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    2nd term = 22 − 3 × 2 + 5 = 4 − 3 × 2 + 5 = 4 − 6 + 5 = 3
  4. Replace n by 3 to find the 3rd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    3rd term = 32 − 3 × 3 + 5 = 9 − 3 × 3 + 5 = 9 − 9 + 5 = 5
  5. Replace n by 4 to find the 4th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    4th term = 42 − 3 × 4 + 5 = 16 − 3 × 4 + 5 = 16 − 12 + 5 = 9
  6. Replace n by 5 to find the 5th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    5th term = 52 − 3 × 5 + 5 = 25 − 3 × 5 + 5 = 25 − 15 + 5 = 15
  7. Write the terms as a list.
    3, 3, 5, 9, 15

Check. The first differences are 0, 2, 4, 6; the second differences are all 2, which is 2 × 1 as expected. ✓

Watch out. n² means n × n, not 2 × n. And the first term is when n = 1, not n = 0.

Watch out

Replace n by the position, then follow BIDMAS. For 2n² with n = 5, square first: 2 × 25 = 50. Do not work out 2 × 5 = 10 and then square it. Also remember that the first term is n = 1, not n = 0.

Exercise 11C  Position-to-term rules

Non-calculator
1
FWrite down the first four terms of the sequence with each nth term.
(a)3n + 1
(b)5n − 2
(c)4n + 3
(d)2n − 7
(e)10n + 5
(f)6n − 1
2
FWrite down the first five terms of the sequence with each nth term.
(a)20 − 3n
(b)12 − 2n
(c)50 − 10n
(d)1 − n
3
FWrite down the first four terms of the sequence with each nth term.
(a)n2 + 1
(b)n2 − 1
(c)2n2
(d)3n2 + 2
(e)n3
(f)3n
4
FWork out the term asked for.
(a)The nth term is 4n + 7. Find the 100th term.
(b)The nth term is 3n − 5. Find the 50th term.
(c)The nth term is 9n + 4. Find the 20th term.
(d)The nth term is 100 − n. Find the 60th term.
(e)The nth term is n2 + 3. Find the 12th term.
(f)The nth term is 2n2 + 1. Find the 10th term.
5
FThe nth term of a sequence is 3n − 2. Work out the terms in positions 1, 2, 5 and 10.
6
PSA taxi charges a fixed £3 plus £2 for every mile. The cost, in pounds, of a journey of n miles is 2n + 3.
(a)Work out the cost of a 7-mile journey.
(b)Work out the cost of a 15-mile journey.
(c)Work out the cost of a 40-mile journey.
7
PSA gym charges £25 to join and then £15 a month. After n months the total cost in pounds is 15n + 25. Work out the total cost after 12 months.
8
RZoe says, “The 5th term of the sequence 2n² is 100, because 2 × 5 = 10 and 10² = 100.” What mistake has she made, and what is the correct answer?
9
RWhich of these is the nth term of the sequence 5, 8, 11, 14, …?   A 5n   B n + 4   C 3n + 2   D 3n + 5. Show how you decide.
10
RThe nth term of one sequence is n² and the nth term of another is 11n.
(a)Which is bigger, the 10th term of n² or the 10th term of 11n?
(b)Which is bigger, the 12th term of each?
11
PSA function machine does the following to the position number n: multiply by 4, then subtract 3, giving the term.
n× 44n− 34n − 3
(a)Write the nth term.
(b)Work out the 6th term.
12
PSThe nth term of a sequence is 7n − 4. Work out the sum of the 1st term and the 10th term.
13
FWrite down the first five terms of the sequence with each nth term.
(a)n2 − 3n + 5
(b)2n2 + n − 1
(c)−n2 + 10n
(d)n2 + n
14
RThe nth term of a sequence is n² − 6n + 10. Show that the 2nd term and the 4th term are equal. Find another pair of equal terms.
11.4

Arithmetic sequences and the nth term

Edexcel 1MA1: A25, A24
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Arithmetic sequences and the nth termComing soon

An arithmetic sequence (also called a linear sequence) goes up or down by the same amount every time. That amount is the common difference, written d. For 5, 8, 11, 14, … the common difference is 3. For 20, 17, 14, 11, … it is −3, because the sequence goes down. We would like a formula for the nth term, so that we can find any term without listing the others. Here is a method that works every time. We will find the nth term of 5, 8, 11, 14, …

Step 1. Find the common difference. 8 − 5 = 3, 11 − 8 = 3, 14 − 11 = 3. So d = 3.

Step 2. Write the times table of d and put it under the sequence. Each time n goes up by 1 the term goes up by 3, just like the 3 times table: 3, 6, 9, 12. So the nth term must start with 3n.

Position n1234
Sequence581114
3n36912
Sequence − 3n2222

Step 3. Compare. Subtract the times table from the sequence: 5 − 3 = 2, 8 − 6 = 2, 11 − 9 = 2, 14 − 12 = 2. The answer is always 2, so we must add 2.

Step 4. Write the nth term: 3n + 2. Check it: when n = 1, 3 × 1 + 2 = 5 ✓; when n = 4, 3 × 4 + 2 = 14 ✓.

Why does it work? The sequence starts at 5. If we went one step back from the first term we would reach 5 − 3 = 2. That “zeroth term” is exactly the number we add. So a quick way to find the adjustment is first term − d. In the picture the terms lie on a straight line, which is why these sequences are called linear. The common difference d is how steep the line is.

nterm1234564812162058111417
The terms of 3n + 2 plotted against position. They lie on a straight line.

For a decreasing sequence the common difference is negative. For 20, 17, 14, 11, … we have d = −3, so the nth term starts with −3n. The times table is −3, −6, −9, −12. The sequence minus this is 23 each time, so the nth term is −3n + 23, which you may also write as 23 − 3n.

Some sequences you should know by heart: the even numbers 2, 4, 6, … have nth term 2n; the odd numbers 1, 3, 5, … have nth term 2n − 1; the multiples of 5 have nth term 5n. Once you have the nth term you can find any term, for example the 50th, by replacing n with 50.

Key ideas

  • An arithmetic sequence has a common difference d (the same amount added each time).
  • The nth term is dn + (adjustment), written like 3n + 2.
  • Method: find d; write the d times table under the sequence; subtract to find the adjustment; write the nth term.
  • The adjustment is also first term − d.
  • Always check the formula with n = 1 and another position.
Worked example 11.4.1
Find the nth term of the sequence 5, 8, 11, 14, …

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Use the common difference to start the nth term (it is the number in front of n), then work out what has to be added or subtracted.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term581114
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    8 − 5 = 3
    11 − 8 = 3
    14 − 11 = 3
    Common difference d = 3
  3. Write the 3 times table underneath. Each time the position goes up by 1, the term changes by 3. That is just like the 3 times table: 3, 6, 9, … So the nth term must start with 3n.
    Position n1234
    Term581114
    3n36912
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term581114
    3n36912
    Term − 3n2222
    The adjustment is always 2. So we must add 2 to 3n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 3n + 2
    Shortcut for later: the adjustment 2 is the term before the first term, 5 − 3 = 2.

Check. Test the formula on terms we already know. n = 1: 3 × 1 + 2 = 3 + 2 = 5 ✓   n = 4: 3 × 4 + 2 = 12 + 2 = 14 ✓

Watch out. The number in front of n is the common difference, not the first term. Check with the first term: if your formula does not give it, something is wrong.

Worked example 11.4.2
Find the nth term of the sequence 4, 9, 14, 19, 24, …

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Use the common difference to start the nth term (it is the number in front of n), then work out what has to be added or subtracted.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n12345
    Term49141924
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    9 − 4 = 5
    14 − 9 = 5
    19 − 14 = 5
    24 − 19 = 5
    Common difference d = 5
  3. Write the 5 times table underneath. Each time the position goes up by 1, the term changes by 5. That is just like the 5 times table: 5, 10, 15, … So the nth term must start with 5n.
    Position n12345
    Term49141924
    5n510152025
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term49141924
    5n510152025
    Term − 5n−1−1−1−1−1
    The adjustment is always −1. So we must subtract 1 from 5n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 5n − 1
    Shortcut for later: the adjustment −1 is the term before the first term, 4 − 5 = −1.

Check. Test the formula on terms we already know. n = 1: 5 × 1 − 1 = 5 − 1 = 4 ✓   n = 5: 5 × 5 − 1 = 25 − 1 = 24 ✓

Watch out. The number in front of n is the common difference, not the first term. Check with the first term: if your formula does not give it, something is wrong.

Worked example 11.4.3
Find the nth term of the sequence 20, 17, 14, 11, … and use it to find the 40th term.

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Use the common difference to start the nth term (it is the number in front of n), then work out what has to be added or subtracted.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term20171411
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    17 − 20 = −3
    14 − 17 = −3
    11 − 14 = −3
    Common difference d = −3
  3. Write the −3 times table underneath. Each time the position goes up by 1, the term changes by −3. That is just like the −3 times table: −3, −6, −9, … So the nth term must start with −3n.
    Position n1234
    Term20171411
    −3n−3−6−9−12
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term20171411
    −3n−3−6−9−12
    Term − −3n23232323
    The adjustment is always 23. So we must add 23 to −3n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = −3n + 23   (you can also write it as 23 − 3n)
    Shortcut for later: the adjustment 23 is the term before the first term, 20 − (−3) = 23.
  6. Replace n by 40 to find the 40th term. The 40th term is the one in position 40.
    40th term = −3 × 40 + 23 = −120 + 23 = −97

Check. Test the formula on terms we already know. n = 1: −3 × 1 + 23 = −3 + 23 = 20 ✓   n = 4: −3 × 4 + 23 = −12 + 23 = 11 ✓

Watch out. This sequence is going down, so the common difference is negative and the number in front of n is negative. Check with the first term.

Worked example 11.4.4
The first term of an arithmetic sequence is 6 and the common difference is 4. Find the nth term.

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. The nth term is (common difference) × n, plus an adjustment so that the first term comes out right.

  1. Write down what we know. The first term is the term when n = 1. The common difference is how much the sequence goes up (or down) each time.
    first term = 6, common difference d = 4
  2. Start the nth term with the common difference times n. Each step changes the term by 4, so the nth term starts with 4n.
    nth term = 4n + (adjustment)
  3. Find the adjustment. When n = 1 the term must be 6. The adjustment is whatever has to be added to 4n to make 6.
    4n with n = 1 gives 4
    adjustment = 6 − 4 = 2
  4. Write the nth term.
    nth term = 4n + 2

Check. The first four terms should be 6, 10, 14, 18, … Test: n = 1: 4 × 1 + 2 = 4 + 2 = 6 ✓   n = 2: 4 × 2 + 2 = 8 + 2 = 10 ✓

Watch out. The number in front of n is the common difference, not the first term.

Worked example 11.4.5
Find the nth term of 7, 12, 17, 22, … Hence find the 50th term.

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Use the common difference to start the nth term (it is the number in front of n), then work out what has to be added or subtracted.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term7121722
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    12 − 7 = 5
    17 − 12 = 5
    22 − 17 = 5
    Common difference d = 5
  3. Write the 5 times table underneath. Each time the position goes up by 1, the term changes by 5. That is just like the 5 times table: 5, 10, 15, … So the nth term must start with 5n.
    Position n1234
    Term7121722
    5n5101520
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term7121722
    5n5101520
    Term − 5n2222
    The adjustment is always 2. So we must add 2 to 5n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 5n + 2
    Shortcut for later: the adjustment 2 is the term before the first term, 7 − 5 = 2.
  6. Replace n by 50 to find the 50th term. The 50th term is the one in position 50.
    50th term = 5 × 50 + 2 = 250 + 2 = 252

Check. Test the formula on terms we already know. n = 1: 5 × 1 + 2 = 5 + 2 = 7 ✓   n = 4: 5 × 4 + 2 = 20 + 2 = 22 ✓

Watch out. The number in front of n is the common difference, not the first term. Check with the first term: if your formula does not give it, something is wrong.

Worked example 11.4.6
Find the nth term of the sequence 32, 2, 52, 3, …

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Use the common difference to start the nth term (it is the number in front of n), then work out what has to be added or subtracted.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term322523
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    2 − 32 = 12
    52 − 2 = 12
    3 − 52 = 12
    Common difference d = 12
  3. Write the multiples of 12 underneath. Each time the position goes up by 1, the term changes by 12. That is just like the multiples of 12: 12, 1, 32, … So the nth term must start with 12n.
    Position n1234
    Term322523
    12n121322
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term322523
    12n121322
    Term − 12n1111
    The adjustment is always 1. So we must add 1 to 12n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 12n + 1
    Shortcut for later: the adjustment 1 is the term before the first term, 32 − 12 = 1.

Check. Test the formula on terms we already know. n = 1: 12 × 1 + 1 = 12 + 1 = 32 ✓   n = 4: 12 × 4 + 1 = 2 + 1 = 3 ✓

Watch out. The number in front of n is the common difference, not the first term. Check with the first term: if your formula does not give it, something is wrong.

Watch out

The number in front of n is the common difference, not the first term. For 5, 8, 11, 14 the nth term is 3n + 2, not 5n + 3. Always test your formula with n = 1: it must give the first term.

Exercise 11D  The nth term of an arithmetic sequence

Non-calculator
1
FFind the nth term of each sequence.
(a)4, 7, 10, 13, …
(b)5, 7, 9, 11, …
(c)3, 8, 13, 18, …
(d)1, 5, 9, 13, …
(e)10, 20, 30, 40, …
(f)2, 9, 16, 23, …
2
FFind the nth term of each decreasing sequence.
(a)20, 17, 14, 11, …
(b)30, 26, 22, 18, …
(c)50, 45, 40, 35, …
(d)9, 7, 5, 3, …
3
FFind the nth term of each sequence, then work out the term asked for.
(a)3, 5, 7, 9, …   Find the 50th term.
(b)6, 11, 16, 21, …   Find the 30th term.
(c)2, 8, 14, 20, …   Find the 100th term.
(d)40, 36, 32, 28, …   Find the 12th term.
4
FFind the nth term of each arithmetic sequence from its first term and common difference.
(a)First term 7, common difference 3
(b)First term 2, common difference 5
(c)First term 20, common difference -4
(d)First term 1, common difference 10
5
RBen says the nth term of 6, 10, 14, 18, … is 6n + 4. Show that he is wrong and find the correct nth term.
6
PSAhmed saves £12 in week 1, £17 in week 2, £22 in week 3, and so on. Each week he saves £5 more than the week before. How much does he save in week 20?
7
PSTables are pushed together in a row. One table seats 6, two tables seat 10, three tables seat 14, and so on. Find the nth term and use it to work out how many people can sit at 25 tables.
8
PSThe cost, in pounds, of hiring a bike for 1, 2, 3, 4 hours is £9, £13, £17, £21. The costs form an arithmetic sequence. Find the nth term and the cost of hiring the bike for 10 hours.
9
RShow that the nth term of the sequence 7, 11, 15, 19, … is 4n + 3.
10
RAli says the nth term of 8, 11, 14, … is 3n + 5. Beth says it is 5 + 3n. Who is right?
11
FThe nth term of an arithmetic sequence is 4n − 1. Write down the first term and the common difference.
12
RShow that every term of the sequence with nth term 3n + 2 is 2 more than a multiple of 3. Use this to explain why 100 is not a term.
13
FFind the nth term of each sequence.
(a)32, 2, 52, 3, …
(b)72, 4, 92, 5, …
(c)3, 72, 4, 92, …
(d)13, 23, 1, 43, …
14
RThe first three terms of an arithmetic sequence are x, 2x + 1 and 4x − 3. Find the value of x and the nth term of the sequence.
11.5

Using the nth term

Edexcel 1MA1: A25, A24
Grade 4-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Using the nth termComing soon

Once you have an nth term you can do much more than find single terms. This section shows four uses.

1. Is a number in the sequence? A number is a term only if it comes from a whole-number position such as 1, 2, 3, … You cannot have the 4.5th term. So set the nth term equal to the number and solve for n. If n is a positive whole number, the number is in the sequence (and n is its position). If not, it is not. For example, is 62 in the sequence 4n − 3? Solving 4n − 3 = 62 gives 4n = 65 and n = 16.25, which is not a whole number, so no.

2. Which term is it? The same method gives the position. If 5n + 2 = 97 then 5n = 95, so n = 19: 97 is the 19th term.

3. The first term bigger than a number. Write an inequality, for example 7n − 4 > 200. Solve it as an equation (7n > 204, so n > 29.14…) and then pick the next whole number up, which is 30. The 30th term is 206, and the 29th is 199, which is too small. In the same way, for the largest term below a number you round down.

4. The nth term of a pattern. To find a formula for a pattern of matchsticks or tiles, count each pattern to make a sequence, find its nth term, and then check that the formula matches how the picture is built. For the rows of squares, the counts 4, 7, 10, 13 have common difference 3 and nth term 3n + 1. In the picture, each square has 3 matchsticks of its own and there is 1 extra at the start.

Pattern 1Pattern 2Pattern 3Pattern 4
The nth term of the matchstick pattern is 3n + 1. Pattern 30 needs 3 × 30 + 1 = 91 matchsticks.

Key ideas

  • A number is in the sequence only if solving “nth term = number” gives a positive whole number n.
  • To find the first term above a value, solve an inequality and round n up; for the largest term below a value, round down.
  • For a pattern, count to make a sequence, find the nth term, then explain it from the picture.
  • Higher: if two terms of an arithmetic sequence are known, divide the change in the terms by the change in the positions to get d.
Worked example 11.5.1
Is 62 a term of the sequence with nth term 4n − 3?

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. A number is in a sequence only if it is the term in a whole-number position. Set the nth term equal to the number and solve for n.

  1. Understand what the question asks. A number is in the sequence only if some position n gives it. Positions are whole numbers: 1, 2, 3, … So we need to find out whether n comes out as a whole number.
    nth term = 4n − 3
    Is there a whole number n with 4n − 3 = 62?
  2. Set the nth term equal to the number and solve for n. Add 3 on both sides to get the n term on its own, then divide by 4.
    4n − 3 = 62
    4n = 62 + 3 = 65
  3. Divide by 4. The inverse of multiplying by 4 is dividing by 4.
    n = 65 ÷ 4 = 16.25
  4. Decide. 16.25 is not a whole number, so there is no such position.
    No: 62 is not in the sequence.

Check. The terms either side are the 16th term, 61, and the 17th term, 65. 62 falls in the gap between them, so it is skipped. ✓

Watch out. Do not just say "it goes up in 4s so it is in the sequence". A number can be close to a term and still not be one. Work out n and check that it is a whole number.

Worked example 11.5.2
Here are the first four terms of a sequence: 7, 12, 17, 22, … Is 97 a term of the sequence? If so, which term?

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. A number is in a sequence only if it is the term in a whole-number position. Set the nth term equal to the number and solve for n.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term7121722
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    12 − 7 = 5
    17 − 12 = 5
    22 − 17 = 5
    Common difference d = 5
  3. Write the 5 times table underneath. Each time the position goes up by 1, the term changes by 5. That is just like the 5 times table: 5, 10, 15, … So the nth term must start with 5n.
    Position n1234
    Term7121722
    5n5101520
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term7121722
    5n5101520
    Term − 5n2222
    The adjustment is always 2. So we must add 2 to 5n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 5n + 2
    Shortcut for later: the adjustment 2 is the term before the first term, 7 − 5 = 2.
  6. Understand what the question asks. A number is in the sequence only if some position n gives it. Positions are whole numbers: 1, 2, 3, … So we need to find out whether n comes out as a whole number.
    nth term = 5n + 2
    Is there a whole number n with 5n + 2 = 97?
  7. Set the nth term equal to the number and solve for n. Subtract 2 on both sides to get the n term on its own, then divide by 5.
    5n + 2 = 97
    5n = 97 − 2 = 95
  8. Divide by 5. The inverse of multiplying by 5 is dividing by 5.
    n = 95 ÷ 5 = 19
  9. Decide. 19 is a whole number and it is at least 1, so it is a real position.
    Yes: 97 is the 19th term.

Check. Substitute n = 19: 5 × 19 + 2 = 95 + 2 = 97, which is 97. ✓

Watch out. Do not just say "it goes up in 4s so it is in the sequence". A number can be close to a term and still not be one. Work out n and check that it is a whole number.

Worked example 11.5.3
The nth term of a sequence is 7n − 4. Find the first term that is greater than 200.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Set up and solve an inequality to find how far along the sequence you must go, then pick the next whole number.

  1. Write what we want as an inequality. We want the first term that is greater than 200. The terms must be greater than 200.
    7n − 4 > 200
  2. Solve it like an equation. Add 4 on both sides, then divide both sides by 7.
    7n > 200 + 4 = 204
    n > 204 ÷ 7 = 2047
  3. Choose the smallest whole number that works. Positions must be whole numbers. n has to be bigger than 2047, so take the next whole number up.
    n = 30
  4. Work out the term and test the one before it. The term in position n must be greater than 200, and the term before it must not be.
    30th term = 7 × 30 − 4 = 210 − 4 = 206
    29th term = 7 × 29 − 4 = 203 − 4 = 199 (not greater than 200)

Check. 206 is greater than 200 and 199 is not. ✓

Watch out. Do not round n to the nearest whole number. You need the first whole number that makes the term big (or small) enough, which means rounding up.

Worked example 11.5.4
The nth term of a sequence is 8n + 3. Find the largest term that is less than 500.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Solve an inequality to find how far along the sequence you can go, then pick the largest whole number that works.

  1. Write what we want as an inequality. We want terms that are less than 500.
    8n + 3 < 500
  2. Solve it like an equation. Subtract 3 on both sides, then divide both sides by 8.
    8n < 497
    n < 497 ÷ 8 = 62.125
  3. Choose the largest whole number that works. n must be a whole number less than 62.125.
    n = 62
  4. Work out the term and test the next one. The term in position n must be less than 500, and the next one must not be.
    62nd term = 8 × 62 + 3 = 496 + 3 = 499
    63rd term = 8 × 63 + 3 = 504 + 3 = 507 (too big)

Check. 499 < 500 but 507 > 500 or equal. ✓

Watch out. Round down here, because the term must stay below the limit.

Worked example 11.5.5
Here are the first three patterns made from matchsticks.
Pattern 1Pattern 2Pattern 3
Find the nth term. How many matchsticks are needed for pattern 30?

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Count the matchsticks, find the rule for the sequence of counts, and use the picture to see why it works.

  1. Count the matchsticks in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
    Pattern number n123
    Number of matchsticks4710
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    7 − 4 = 3
    10 − 7 = 3
    Common difference d = 3
  3. Write the 3 times table underneath. Each time the position goes up by 1, the term changes by 3. That is just like the 3 times table: 3, 6, 9, … So the nth term must start with 3n.
    Position n123
    Term4710
    3n369
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term4710
    3n369
    Term − 3n111
    The adjustment is always 1. So we must add 1 to 3n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 3n + 1
    Shortcut for later: the adjustment 1 is the term before the first term, 4 − 3 = 1.
  6. Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
    The first square needs 4 matchsticks. Each new square is added on the end and shares one side with the square before it, so it needs only 3 new matchsticks. After n squares there are 3 matchsticks for each square, plus 1 extra for the very first side: 3n + 1.
  7. Replace n by 30. Pattern 30 is the one in position 30.
    Pattern 30: 3 × 30 + 1 = 90 + 1 = 91

Check. Test the rule on patterns you can see. n = 1: 3 × 1 + 1 = 3 + 1 = 4 ✓   n = 3: 3 × 3 + 1 = 9 + 1 = 10 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

Worked example 11.5.6
The 3rd term of an arithmetic sequence is 11 and the 7th term is 27. Find the nth term.

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Work out the common difference from the two terms (change in terms ÷ change in positions), then find what must be added.

  1. Find how many steps there are between the two terms. Going from the 3rd term to the 7th term means taking steps of the common difference d. The number of steps is the difference between the positions.
    7 − 3 = 4 steps
  2. Find the total change in the terms. This is how much the sequence has gone up (or down) over those steps.
    27 − 11 = 16
  3. Divide to find the common difference. 4 equal steps make a total change of 16, so one step is the total divided by 4.
    d = 16 ÷ 4 = 4
  4. Use the common difference to start the nth term. The nth term of an arithmetic sequence starts with dn.
    nth term = 4n + (something)
  5. Put the 3rd term into the formula to find the something. We know that when n = 3 the term is 11. Let the missing number be c.
    4n + c with n = 3: 4 × 3 + c = 11
    12 + c = 11
    c = 11 − 12 = −1
  6. Write the nth term.
    nth term = 4n − 1

Check. Test both terms. n = 3: 4 × 3 − 1 = 12 − 1 = 11 ✓   n = 7: 4 × 7 − 1 = 28 − 1 = 27 ✓

Watch out. The common difference is the change in the terms divided by the change in the positions, not divided by the change in the terms.

Worked example 11.5.7
The nth term of one sequence is 3n + 2 and the nth term of another is 5n − 6. Is there a position where the two sequences have the same term?

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Make the two nth terms equal and solve for n.

  1. Say what the question means. The two sequences have the same term in the same position n when their nth terms are equal. Write the one with the bigger number of ns first.
    5n − 6 = 3n + 2
  2. Subtract 3n from both sides. This gets all the n terms onto one side. Doing the same to both sides keeps the equation balanced.
    2n − 6 = 2
  3. Add 6 on both sides. This leaves the n term on its own.
    2n = 8
  4. Divide both sides by 2.
    n = 8 ÷ 2 = 4
  5. Check that n is a position. A position must be a positive whole number. It is.
    n = 4. Both sequences have the term 14 in position 4.

Check. Substitute: 3 × 4 + 2 = 12 + 2 = 14 and 5 × 4 − 6 = 20 − 6 = 14. ✓

Watch out. Subtract the smaller number of ns from both sides so that you keep a positive number of ns.

Watch out

Do not round n to the nearest whole number. If 7n − 4 > 200 gives n > 29.14, then n = 29 does not work (the 29th term is 199). You need the next whole number up, 30. And a number is only in the sequence if n comes out as an exact whole number.

Exercise 11E  Using the nth term

Non-calculator
1
FThe nth term of a sequence is 3n + 1. Is each number a term of the sequence? If it is, say which term.
(a)31
(b)50
(c)100
(d)151
2
FIs each number a term of the sequence given? Show how you decide.
(a)Is 100 in 7, 11, 15, 19, …
(b)Is 99 in 3, 9, 15, 21, …
(c)Is 61 in 2, 5, 8, 11, …
(d)Is 87 in 10, 15, 20, 25, …
3
FThe nth term of a sequence is 20 − 3n. Is each number a term of the sequence?
(a)-10
(b)-11
(c)2
(d)0
4
FFind the first term of each sequence that is greater than the given number.
(a)nth term 7n − 4; greater than 100
(b)nth term 4n + 1; greater than 250
(c)nth term 9n + 2; greater than 500
(d)nth term 2n + 5; greater than 1000
5
FThe nth term of each sequence is given. Find the first term that is less than 0.
(a)nth term 50 − 3n
(b)nth term 42 − 5n
(c)nth term 100 − 7n
6
FHere are the first three patterns made from matchsticks.
Pattern 1Pattern 2Pattern 3
(a)Find the nth term.
(b)How many matchsticks are needed for pattern 50?
7
PSHere are the first three patterns of triangles made from matchsticks.
Pattern 1Pattern 2Pattern 3
(a)Find the nth term.
(b)Which pattern uses exactly 71 matchsticks?
8
PSHere are the first three tile patterns.
Pattern 1Pattern 2Pattern 3
(a)Find the nth term.
(b)How many tiles are in pattern 20?
(c)Which pattern has 101 tiles?
9
PSA square pond is surrounded by a border of paving slabs. Here are the first three patterns, with the pond in blue and the slabs in orange.
Pattern 1Pattern 2Pattern 3
(a)Find the nth term for the number of slabs.
(b)How many slabs are needed to go round a pond that is 25 slabs wide (pattern 25)?
10
PSA taxi company charges £3 plus £2 for each mile. The cost, in pounds, of a journey of n miles is 2n + 3. Could a journey cost exactly the amount below? If so, how many miles was it?
(a)£47
(b)£50
11
PSTables are pushed together in a row. A row of n tables seats 4n + 2 people. A club wants every seat to be taken. Could it be done with exactly this many people? If so, how many tables?
(a)100 people
(b)102 people
12
PSThe nth term of a sequence is 11n + 5. Find the largest term of the sequence that is less than 300.
13
RThe nth term of a sequence is 6n − 2. Explain why none of the terms is a multiple of 3.
14
FFind the nth term of each arithmetic sequence from the two terms given.
(a)The 3rd term is 11 and the 7th term is 27
(b)The 2nd term is 5 and the 6th term is 21
(c)The 4th term is 20 and the 9th term is 5
(d)The 5th term is 12 and the 8th term is 21
15
RFor each pair of sequences, find the position where they have the same term, or show that there is no such position.
(a)5n + 3 and 8n − 12
(b)4n + 1 and 6n − 5
(c)3n + 7 and 5n + 2
16
RThe nth term of a sequence is 2n + 1. Explain why every term is an odd number. Hence show that no pattern of triangles made from matchsticks (3, 5, 7, 9, … matchsticks) can use exactly 100 matchsticks.
11.6

Special sequences

Edexcel 1MA1: A24
Grade 2-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Special sequencesComing soon

Some sequences turn up so often that they have names. You should know them well: you will see them in non-calculator exams. Here they are, with the nth term of each.

Square numbers14916253649…
Cube numbers182764125216343…
Triangular numbers13610152128…
Powers of 2248163264128…
Powers of 1010100100010 000…

Square numbers. The nth term is n² = n × n. They are the numbers of dots in a square pattern. Cube numbers. The nth term is n³ = n × n × n. They count the small cubes in a bigger cube.

Pattern 1Pattern 2Pattern 3Pattern 4
Square numbers: 1, 4, 9, 16.
1³ = 12³ = 83³ = 27
Cube numbers: 1, 8, 27.

Triangular numbers. These count the dots in triangles with rows of 1, 2, 3, … dots: 1, 3, 6, 10, 15, … The nth term is n(n + 1) ÷ 2. Why? Put two identical triangles together and you get a rectangle that is n dots by (n + 1) dots. One triangle is half of the rectangle. Here is the idea for n = 4: two triangles of 10 dots make a 4 by 5 rectangle of 20 dots.

Pattern 1Pattern 2Pattern 3Pattern 4
Triangular numbers: 1, 3, 6, 10. The differences are 2, 3, 4, …

Powers of 2 and 10. These are made by multiplying by 2 (or by 10) each time. They are examples of geometric sequences (Section 11.7). The nth term is 2n, which means n twos multiplied together, so 25 = 2 × 2 × 2 × 2 × 2 = 32.

Fibonacci-type sequences. In a Fibonacci-type sequence you make each term by adding the two terms before it. The famous Fibonacci sequence starts 1, 1, 2, 3, 5, 8, 13, 21, … Other Fibonacci-type sequences start with different numbers, for example 2, 5, 7, 12, 19, … (2 + 5 = 7, 5 + 7 = 12, 7 + 12 = 19). To find a missing term, add the two terms before it, or work backwards by subtracting.

How to recognise them. Look at the differences and ratios. Differences 3, 5, 7, 9 (odd numbers going up by 2) mean square numbers. Differences 2, 3, 4, 5 mean triangular numbers. If each term is double the last, they are powers of 2. If each term is the sum of the two before, it is Fibonacci-type.

Key ideas

  • Squares: 1, 4, 9, 16, 25, …   nth term n²
  • Cubes: 1, 8, 27, 64, 125, …   nth term n³
  • Triangular: 1, 3, 6, 10, 15, …   nth term n(n + 1) ÷ 2
  • Powers of 2: 2, 4, 8, 16, 32, …   nth term 2n
  • Fibonacci-type: each term is the sum of the two before it.
  • Learn the first ten squares, the first five cubes and the first eight triangular numbers.
Worked example 11.6.1
Work out the 15th triangular number.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Use the nth-term rule for triangular numbers.

  1. Remember what triangular numbers are. A triangular number is the number of dots in a triangle with rows of 1, 2, 3, … dots. The nth triangular number is 1 + 2 + … + n. Two copies of the same triangle fit together to make a rectangle n by (n + 1), so one copy is half of that: n(n + 1) ÷ 2.
    nth term = n(n + 1)2
  2. Replace n by 15. We want the term in position 15, so use 15 in place of every n.
    15th term = 15 × (15 + 1) ÷ 2
  3. Work it out. Follow the order of operations: brackets, powers, multiply and divide, then add and subtract.
    15th term = 15 × (15 + 1) ÷ 2 = 15 × 16 ÷ 2 = 240 ÷ 2 = 120

Check. Add the numbers 1 to 15 directly: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120. ✓

Watch out. Multiply n by (n + 1) first, and only then divide by 2.

Worked example 11.6.2
Write down the name of the sequence 1, 8, 27, 64, … and the next two terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Recognise the sequence by its differences or ratios and by comparing with the special sequences you know.

  1. Find the differences between neighbouring terms. This tells us what kind of sequence it is.
    8 − 1 = 7
    27 − 8 = 19
    64 − 27 = 37
    These are 7, 19, 37: going up by a bigger amount each time (the differences 7, 19, 37, … are not simple).
  2. Compare with the special sequences you know. It helps to know these by heart: squares 1, 4, 9, 16, 25; cubes 1, 8, 27, 64, 125; triangular numbers 1, 3, 6, 10, 15; powers of 2: 2, 4, 8, 16; powers of 10: 10, 100, 1000.
    The cube numbers are 1, 8, 27, 64, 125, 216, 343, …
    Our sequence 1, 8, 27, 64 appears in this list.
  3. Use the rule for the cube numbers to find the next 2 terms. The rule is nth term = n3.
    5th term of the cube numbers: 53 = 125
    6th term of the cube numbers: 63 = 216

Check. The next terms 125, 216 carry on the same pattern as 27, 64. ✓

Watch out. A sequence that looks like squares might be something else. Check at least three terms, and use the differences.

Worked example 11.6.3
The first two terms of a Fibonacci-type sequence are 2 and 3. Write down the first eight terms.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. A Fibonacci-type sequence is built by adding the previous two terms.

  1. Write the first two terms. A Fibonacci-type sequence needs two starting terms.
    2, 3
  2. Add the last two terms to make each new term. In a Fibonacci-type sequence every term is the sum of the two terms before it.
    Term 3 = 2 + 3 = 5
    Term 4 = 3 + 5 = 8
    Term 5 = 5 + 8 = 13
    Term 6 = 8 + 13 = 21
    Term 7 = 13 + 21 = 34
    Term 8 = 21 + 34 = 55
  3. Write the sequence.
    2, 3, 5, 8, 13, 21, 34, 55

Check. Check the last term: 21 + 34 = 55. ✓

Watch out. Always add the last two terms, not the first two.

Worked example 11.6.4
Here is a Fibonacci-type sequence: 3, 5,  ? , 13,  ?  Find the missing terms.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Every term is the sum of the previous two. To find a missing term, add the two before it, or subtract to work backwards.

  1. Use the rule: each term is the sum of the two terms before it. Look for a gap where you know the two terms before it (add them) or the two terms after it (subtract).
    Terms: 3, 5, ?, 13, ?
  2. Find term 3. Term 3 is the sum of the two terms before it.
    3 + 5 = 8
  3. Find term 5. Term 5 is the sum of the two terms before it.
    8 + 13 = 21
  4. Write the complete sequence.
    3, 5, 8, 13, 21

Check. Check every term is the sum of the two before it: 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21. ✓

Watch out. When working backwards, subtract: the missing term is the later term minus its neighbour.

Worked example 11.6.5
Here is a Fibonacci-type sequence:  ? ,  ? , 7, 11, 18. Find the first two terms.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Every term is the sum of the previous two. To find a missing term, add the two before it, or subtract to work backwards.

  1. Use the rule: each term is the sum of the two terms before it. Look for a gap where you know the two terms before it (add them) or the two terms after it (subtract).
    Terms: ?, ?, 7, 11, 18
  2. Find term 2. Term 4 is the sum of terms 2 and 3, so term 2 = term 4 − term 3. We are working backwards.
    11 − 7 = 4
  3. Find term 1. Term 3 is the sum of terms 1 and 2, so term 1 = term 3 − term 2. We are working backwards.
    7 − 4 = 3
  4. Write the complete sequence.
    3, 4, 7, 11, 18

Check. Check every term is the sum of the two before it: 3 + 4 = 7, 4 + 7 = 11, 7 + 11 = 18. ✓

Watch out. When working backwards, subtract: the missing term is the later term minus its neighbour.

Worked example 11.6.6
Which term of the sequence of square numbers is 144?

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Work backwards from the term to the position.

  1. Remember the rule for square numbers. A square number is a whole number multiplied by itself. It is the number of small squares in a bigger square with that side length.
    nth term = n2
  2. Use the square root. The inverse (undo) of squaring is the square root. We need the number that multiplies by itself to give 144.
    √144 = 12, because 12 × 12 = 144
  3. Write the answer. The position is the value of n.
    144 is the 12th term of the square numbers.

Check. Substitute n = 12: 122 = 144. ✓

Watch out. The answer is the position (n), not the term. Say which one you are giving.

Watch out

n² is n × n, not 2 × n. The square numbers go 1, 4, 9, 16, but 2n goes 2, 4, 6, 8. In the same way, 2n (2, 4, 8, 16, …) is not the same as 2n or n². They only look alike at the start: all three have 4 as the second term.

Exercise 11F  Special sequences

Non-calculator
1
FWork out each of these.
(a)The 9th square number
(b)The 4th cube number
(c)The 7th triangular number
(d)The 8th power of 2
(e)The 15th square number
(f)The 6th cube number
2
FWrite down the name of each sequence and the next two terms.
(a)1, 4, 9, 16, …
(b)1, 8, 27, 64, …
(c)3, 6, 10, 15, …
(d)4, 9, 16, 25, …
(e)2, 4, 8, 16, …
(f)10, 100, 1000, …
3
FWhich term of the sequence is each number?
(a)169 in the square numbers
(b)343 in the cube numbers
(c)55 in the triangular numbers
(d)64 in the powers of 2
(e)1000 in the powers of 10
4
FThe first two terms of each Fibonacci-type sequence are given. Write down the first eight terms.
(a)1, 1, …
(b)2, 3, …
(c)1, 3, …
(d)5, 2, …
5
FEach of these is part of a Fibonacci-type sequence. Find the missing terms.
(a)3 , 5 ,  ?  , 13 ,  ? 
(b) ?  ,  ?  , 7 , 11 , 18
(c)1 ,  ?  ,  ?  , 7 , 11
(d)2 , 5 ,  ?  , 12 ,  ? 
(e) ?  ,  ?  ,  ?  , 11 , 18
6
PSTins are stacked in a triangular pile with 1 tin in the top row, 2 in the next row, 3 in the next row, and so on.
(a)How many tins are in a pile with 9 rows?
(b)A pile has 66 tins. How many rows does it have?
7
PSA cube is made from small cubes. A 5 by 5 by 5 cube is made from 125 small cubes. How many more small cubes are needed to make a 6 by 6 by 6 cube?
8
PSIn a computer game, level 1 needs 2 points, level 2 needs 4 points, level 3 needs 8 points and the number of points doubles with each level. How many points are needed for level 10?
9
RWork out the sum of each pair of consecutive triangular numbers. What do you notice?
(a)3 + 6
(b)6 + 10
(c)10 + 15
10
RIs 50 a triangular number? Give a reason.
11
RPete says that 2n and n² are the same. Show that Pete is wrong.
12
RThe first term of a Fibonacci-type sequence is x and the second term is 3. The sixth term is 36. Find the value of x.
13
RShow that no square number ends in 2, 3, 7 or 8.
11.7

Geometric sequences

Edexcel 1MA1: A24
Grade 3-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Geometric sequencesComing soon

An arithmetic sequence is made by adding the same number each time. A geometric sequence is made by multiplying by the same number each time. That number is the common ratio, written r. For 3, 6, 12, 24, … every term is double the one before, so r = 2. For 81, 27, 9, 3, … each term is a third of the one before, so r = 13 (we can say “divide by 3”). The ratio can also be a decimal: 8, 12, 18, 27 has r = 1.5.

Finding the common ratio. Divide any term by the term before it: 12 ÷ 6 = 2. Check with another pair, because the ratio must be the same everywhere. Geometric sequences grow (or shrink) very fast, much faster than arithmetic ones. Doubling 2, 4, 8, 16, 32, 64, 128 reaches 128 in only seven terms.

nterm123456163248642481632
The terms of 2n plotted against position. They climb in a curve, not a straight line.

Geometric sequences appear in real life: bacteria that double every hour, a ball that rebounds to half its previous height, a phone that loses 20% of its value each year (multiply by 0.8). Money growing with compound interest (Chapter 5) is also geometric.

Position-to-term rules. For the powers of a number the nth term is rn. For example 3n gives 3, 9, 27, 81, … and (12)n gives 12, 14, 18, …. Take care: 3n means 3 multiplied by itself n times, not 3 × n.

Higher: the nth term of any geometric sequence. Suppose the first term is a and the common ratio is r. The 1st term is a. The 2nd term is a × r (multiply once). The 3rd term is a × r × r = ar² (multiply twice). The number of times we multiply is always one less than the position, so the nth term is a × rn − 1. If you know two terms you can find r: going from the 2nd term to the 5th term means multiplying by r three times, so r3 = (5th term) ÷ (2nd term).

Higher: surd ratios. Sometimes the common ratio is a surd, such as √3. The key fact is √3 × √3 = 3. For the sequence 2, 2√3, 6, 6√3, … the ratio is √3: 2 × √3 = 2√3, and 2√3 × √3 = 2 × 3 = 6. Every second term is a whole number.

Key ideas

  • A geometric sequence multiplies by the same common ratio r each time. Find r by dividing a term by the one before it.
  • Divide by a number is the same as multiplying by a fraction: “divide by 2” means r = 12.
  • The powers of a number have nth term rn: 2, 4, 8, 16, … is 2n.
  • Higher: the nth term is a × rn − 1.
  • Higher: √k × √k = k.
Worked example 11.7.1
Write down the next two terms of the sequence 2, 6, 18, 54, …

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 2 terms.

  1. Try the differences first. We always look for "add or subtract the same number" before anything else.
    6 − 2 = 4
    18 − 6 = 12
    54 − 18 = 36
    The differences are not all the same, so it is not an add/subtract rule.
  2. Now divide each term by the term before it. If we are multiplying by the same number each time, dividing a term by the one before gives that number.
    6 ÷ 2 = 3
    18 ÷ 6 = 3
    54 ÷ 18 = 3
  3. Look at the answers. They are all the same, so we multiply by that number each time.
    The rule is multiply by 3.
  4. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    54 × 3 = 162
    162 × 3 = 486
  5. Write down the new terms.
    162, 486

Check. Look back along the whole list: 18, 54, 162, 486. Every term is the one before multiplied by 3. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Worked example 11.7.2
Write down the next two terms of the sequence 80, 40, 20, 10, …

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 2 terms.

  1. Try the differences first. We always look for "add or subtract the same number" before anything else.
    40 − 80 = −40
    20 − 40 = −20
    10 − 20 = −10
    The differences are not all the same, so it is not an add/subtract rule.
  2. Now divide each term by the term before it. If we are multiplying by the same number each time, dividing a term by the one before gives that number.
    40 ÷ 80 = 0.5
    20 ÷ 40 = 0.5
    10 ÷ 20 = 0.5
  3. Look at the answers. They are all the same, so we multiply by that number each time.
    The rule is divide by 2.
  4. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    10 ÷ 2 = 5
    5 ÷ 2 = 2.5
  5. Write down the new terms.
    5, 2.5

Check. Look back along the whole list: 20, 10, 5, 2.5. Every term is the one before divided by 2. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Worked example 11.7.3
Here is a sequence: 8, 12, 18, 27, … Show that it is geometric and find the next term.

Key words. A sequence is a list of numbers that follow a rule. Each number is a term. The term-to-term rule tells you how to get from one term to the next.

Method. Find the rule that takes you from one term to the next, then use it to find the next 1 term.

  1. Try the differences first. We always look for "add or subtract the same number" before anything else.
    12 − 8 = 4
    18 − 12 = 6
    27 − 18 = 9
    The differences are not all the same, so it is not an add/subtract rule.
  2. Now divide each term by the term before it. If we are multiplying by the same number each time, dividing a term by the one before gives that number.
    12 ÷ 8 = 1.5
    18 ÷ 12 = 1.5
    27 ÷ 18 = 1.5
  3. Look at the answers. They are all the same, so we multiply by that number each time.
    The rule is multiply by 1.5.
  4. Use the rule to carry on the sequence. Apply the rule to the last term you have, then to the new term, and so on.
    27 × 1.5 = 40.5
  5. Write down the new terms.
    40.5

Check. Look back along the whole list: 12, 18, 27, 40.5. Every term is the one before multiplied by 1.5. ✓

Watch out. Work out the rule from the terms you are given, and use it on the last term. Do not start again from the first term.

Worked example 11.7.4
The nth term of a sequence is 3n. Write down the first five terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the formula by each position number, and work out the answer.

  1. Write the rule and say what it means. The letter n stands for the position of the term (1 for the first term, 2 for the second, and so on). To get a term, replace n by the position number.
    nth term = 3n
  2. Replace n by 1 to find the 1st term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    1st term = 31 = 3
  3. Replace n by 2 to find the 2nd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    2nd term = 32 = 9
  4. Replace n by 3 to find the 3rd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    3rd term = 33 = 27
  5. Replace n by 4 to find the 4th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    4th term = 34 = 81
  6. Replace n by 5 to find the 5th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    5th term = 35 = 243
  7. Write the terms as a list.
    3, 9, 27, 81, 243

Check. Re-do the last one on a calculator: the 5th term should be 243. ✓

Watch out. n² means n × n, not 2 × n. And the first term is when n = 1, not n = 0.

Worked example 11.7.5
The nth term of a sequence is (12)n. Write down the first four terms.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Replace n in the formula by each position number, and work out the answer.

  1. Write the rule and say what it means. The letter n stands for the position of the term (1 for the first term, 2 for the second, and so on). To get a term, replace n by the position number.
    nth term = (12)n
  2. Replace n by 1 to find the 1st term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    1st term = (12)1 = 12
  3. Replace n by 2 to find the 2nd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    2nd term = (12)2 = 14
  4. Replace n by 3 to find the 3rd term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    3rd term = (12)3 = 18
  5. Replace n by 4 to find the 4th term. Work out powers first, then multiplications, then add or subtract (BIDMAS).
    4th term = (12)4 = 116
  6. Write the terms as a list.
    12, 14, 18, 116

Check. Re-do the last one on a calculator: the 4th term should be 116. ✓

Watch out. n² means n × n, not 2 × n. And the first term is when n = 1, not n = 0.

Worked example 11.7.6
Find the nth term of the geometric sequence 5, 10, 20, 40, …

Key words. A geometric sequence is made by multiplying by the same number every time. That number is the common ratio, written r. You find it by dividing any term by the term before it.

Method. For a geometric sequence the nth term is (first term) × (common ratio)n − 1.

  1. Find the common ratio r. Divide each term by the term before it. In a geometric sequence the answer is always the same.
    10 ÷ 5 = 2
    20 ÷ 10 = 2
    40 ÷ 20 = 2
    r = 2
  2. Say how each term is built from the first term. The first term is 5. To get to term 2 we multiply by r once; to get to term 3 we multiply by r twice; and so on. The number of times we multiply is always one less than the position.
    Term 1 = 5
    Term 2 = 5 × 2 (multiply by r once)
    Term 3 = 5 × 22 (multiply by r 2 times) = 20
    Term 4 = 5 × 23 (multiply by r 3 times) = 40
  3. Write the nth term. To reach position n we multiply the first term by r exactly (n − 1) times.
    nth term = a × rn − 1 = 5 × 2n − 1

Check. Test the formula. n = 1: 5 × 21 − 1 = 5 × 20 = 5 × 1 = 5 ✓   n = 4: 5 × 24 − 1 = 5 × 23 = 5 × 8 = 40 ✓

Watch out. The power is n − 1, not n, because the first term has not been multiplied at all. Check that your formula gives the first term when n = 1.

Worked example 11.7.7
The 2nd term of a geometric sequence is 6 and the 5th term is 162. Find the common ratio and the nth term.

Key words. A geometric sequence is made by multiplying by the same number every time. That number is the common ratio, written r. You find it by dividing any term by the term before it.

Method. The ratio between two terms is r raised to the number of steps between them. Divide, then take a root.

  1. Count the steps from the 2nd term to the 5th term. Each step multiplies by the common ratio r.
    5 − 2 = 3 steps, so 162 = 6 × r3
  2. Divide to isolate the power of r. Divide both sides by 6.
    r3 = 162 ÷ 6 = 27
  3. Take the cube root. We need the positive number that multiplies by itself 3 times to give 27. The inverse of raising to the power 3 is the cube root.
    r = 33 = 27, so r = 3
  4. Go back to the first term. The 2nd term has been multiplied by r exactly 1 time from the first term, so divide the 2nd term by r that many times.
    first term = 6 ÷ 3 = 2
  5. Write the nth term. Use first term × rn − 1.
    nth term = 2 × 3n − 1

Check. Test both given terms. n = 2: 2 × 32 − 1 = 2 × 31 = 2 × 3 = 6 ✓   n = 5: 2 × 35 − 1 = 2 × 34 = 2 × 81 = 162 ✓

Watch out. The number of steps is the difference of the positions, not the difference of the terms.

Worked example 11.7.8
Here is a geometric sequence: 2, 2√3, 6, 6√3, … Find the next two terms.

Key words. A surd is a root that does not work out exactly, such as √2 or √3. The key fact is √k × √k = k. For example √3 × √3 = 3, and 2√3 × √3 = 2 × 3 = 6.

Method. Find the common ratio by dividing, then multiply by it. When two matching surds multiply, the surd disappears: √k × √k = k.

  1. Divide the second term by the first to find the common ratio. In a geometric sequence every term divided by the one before gives the same answer, the common ratio r.
    2√3 ÷ 2 = √3

    62√3: there is a surd on the bottom, so multiply the top and the bottom by √3 (this does not change the value).
    6 × √32√3 × √3 = 6√36 = √3
    r = √3
  2. Check the ratio with the next pair. The ratio must be the same everywhere, so test it on the other terms.
    2√3 ÷ 2 = √3
    6 ÷ 2√3 = √3
    6√3 ÷ 6 = √3
  3. Multiply the last term by r to carry on. Use √k × √k = k whenever two matching surds multiply.
    6√3 × √3 = 6 × 1 × (√3 × √3) = 6 × 3 = 18

    18 × √3 = 18√3
  4. Write down the new terms.
    18, 18√3

Check. Look at every second term: 6√3, 18√3. They should be 3 times apart: 6√3 × 3 = 18√3. ✓

Watch out. Do not write √3 × √3 as √6 or as 2√3. It is exactly 3.

Watch out

Do not mix up the two kinds of sequence. 2, 4, 8, 16 is geometric (multiply by 2), but 2, 4, 6, 8 is arithmetic (add 2). Test the differences first; if they are not the same, test the ratios. And remember that “divide by 2” is a geometric rule with r = 12, even though the numbers go down.

Exercise 11G  Geometric sequences

Non-calculator
1
FWrite down the next two terms of each sequence.
(a)2, 6, 18, 54, …
(b)1000, 100, 10, 1, …
(c)4, 12, 36, 108, …
(d)6, 3, 1.5, …
(e)8, 12, 18, 27, …
(f)2, 10, 50, …
2
FWrite down the common ratio of each geometric sequence.
(a)3, 6, 12, 24, …
(b)2, 10, 50, …
(c)162, 54, 18, 6, …
(d)4, 6, 9, …
(e)1, 4, 16, 64, …
(f)64, 16, 4, 1, …
3
FSay whether each sequence is arithmetic or geometric, and describe the rule.
(a)4, 8, 12, 16, …
(b)4, 8, 16, 32, …
(c)90, 30, 10, …
(d)90, 60, 30, 0, …
(e)7, 21, 63, …
(f)7, 14, 21, 28, …
4
FWrite down the first four terms of the sequence with each nth term.
(a)2n
(b)5n
(c)10n
(d)(12)n
5
FWork out the term asked for.
(a)nth term 2n: the 8th term
(b)nth term 3n: the 6th term
(c)nth term 10n: the 5th term
(d)nth term (12)n: the 4th term
6
PSA scientist starts with 50 bacteria. The number doubles every hour. Term 1 is the start, term 2 is after 1 hour, and so on. How many bacteria are there after 6 hours?
7
PSA ball is dropped from a height of 160 cm. Each time it bounces back up to half its previous height. How high does it bounce on the 4th bounce?
8
PSA phone is worth £500 when new. After 1 year it is worth £400, after 2 years £320. Each year its value is multiplied by the same number. What is its value after 4 years?
9
RLiam says, “The sequence 1, 2, 4, 7, 11 is geometric because it goes up so fast.” Is he right? Explain.
10
RShow that the sequence 3, 12, 48, 192, … is geometric and find the 7th term.
11
RThe nth term of one sequence is 2n and the nth term of another sequence is 2n. Compare the 10th terms of the two sequences. What does this show about geometric sequences?
12
FFind the nth term of each geometric sequence.
(a)5, 10, 20, 40, …
(b)3, 6, 12, 24, …
(c)81, 27, 9, 3, …
(d)16, 24, 36, 54, …
13
FFind the nth term of the geometric sequence, using the two terms given.
(a)The 2nd term is 6 and the 5th term is 162
(b)The 1st term is 4 and the 3rd term is 36
(c)The 3rd term is 12 and the 4th term is 6
(d)The 2nd term is 20 and the 4th term is 5
14
FEach sequence is geometric with a surd as the common ratio. Find the next two terms.
(a)2, 2√3, 6, 6√3, …
(b)1, √2, 2, 2√2, …
(c)√3, 3, 3√3, …
(d)2√3, 6, 6√3, 18, …
(e)5, 5√5, …
15
FWrite down the first five terms of the sequence with nth term (√2)n.
16
RThe first term of a geometric sequence is 3 and the third term is 12. All the terms are positive. Find the common ratio, and then the 5th term.
17
RThe numbers 4, x and 9 are the first three terms of a geometric sequence. Show that x = 6 (given that x is positive).
11.8

Quadratic sequences

Edexcel 1MA1: A24, A25
Grade 5-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Quadratic sequencesComing soon

In an arithmetic sequence the differences between terms are all the same. Look at 3, 8, 15, 24, 35. The first differences are 5, 7, 9, 11. They are not the same, so this is not an arithmetic sequence. But they change in a steady way: each is 2 more than the one before. If we find the differences of the differences we get 2, 2, 2. These are the second differences.

Terms381524351st differences579112nd differences222
Terms (dark), first differences (orange) and second differences (green). The second differences are all 2.

A sequence whose second differences are constant is a quadratic sequence. Its nth term has an n² in it and looks like an² + bn + c. Its graph is a curve, not a straight line.

nterm123456102030405038152435
The terms of n² + 2n. They lie on a curve.

Why the second difference gives a. The square numbers 1, 4, 9, 16, 25 have first differences 3, 5, 7, 9 and second differences 2, 2, 2. So the sequence n² has second difference 2. If every term is multiplied by a, the differences are all multiplied by a too, so an² has second difference 2a. Adding a linear part bn + c does not change the second differences (a linear sequence has constant first differences, so its second differences are 0). Therefore a = (second difference) ÷ 2.

The method.

  1. Write the terms and find the first differences, then the second differences. Check the second differences are all the same.
  2. Halve the second difference to get a.
  3. Work out an² for each position (that is a × 1, a × 4, a × 9, …) and subtract it from each term.
  4. What is left is an arithmetic sequence. Find its nth term, bn + c, using the common difference method from Section 11.4.
  5. Write an² + bn + c and check it with two terms.

If a is negative the sequence curves downwards. If the second difference is odd then a is a fraction such as 12. Quadratic sequences also come from patterns where the number of dots or tiles is made from a square and a few extras.

Key ideas

  • Second differences constant: the sequence is quadratic, with nth term an² + bn + c.
  • a = (second difference) ÷ 2.
  • Subtract an² from every term; the leftover is linear and gives bn + c.
  • Always check the formula on at least two given terms.
  • To test whether a number is a term, set the nth term equal to it and solve for n; n must be a positive whole number.
Worked example 11.8.1
Find the nth term of the quadratic sequence 3, 8, 15, 24, 35, …

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

  1. Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
    Position n12345
    Term38152435
    First differences: 5, 7, 9, 11
    They are not all the same, so it is not a linear sequence.
  2. Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
    Terms381524351st differences579112nd differences222
    Second differences: 2, 2, 2
    They are all 2, so the sequence is quadratic.
  3. Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
    a = 2 ÷ 2 = 1
    The nth term starts with n2.
  4. Subtract n2 from each term. This removes the quadratic part. Work out n2 for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
    Position n12345
    n²1491625
    n21491625
    Term38152435
    Term − n2246810
  5. Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
    Leftover sequence: 2, 4, 6, 8, 10
    Common difference = 4 − 2 = 2, so it starts with 2n.
    Adjustment: 2 − 2 = 0
    Leftover nth term = 2n
  6. Put the two parts together. The nth term is the quadratic part plus the linear part.
    nth term = n2 + (2n) = n2 + 2n

Check. Test the formula. n = 2: 22 + 2 × 2 = 4 + 2 × 2 = 4 + 4 = 8 ✓   n = 5: 52 + 2 × 5 = 25 + 2 × 5 = 25 + 10 = 35 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.2
Find the nth term of the sequence 6, 15, 28, 45, 66, …

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

  1. Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
    Position n12345
    Term615284566
    First differences: 9, 13, 17, 21
    They are not all the same, so it is not a linear sequence.
  2. Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
    Terms6152845661st differences91317212nd differences444
    Second differences: 4, 4, 4
    They are all 4, so the sequence is quadratic.
  3. Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
    a = 4 ÷ 2 = 2
    The nth term starts with 2n2.
  4. Subtract 2n2 from each term. This removes the quadratic part. Work out 2n2 for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
    Position n12345
    n²1491625
    2n228183250
    Term615284566
    Term − 2n247101316
  5. Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
    Leftover sequence: 4, 7, 10, 13, 16
    Common difference = 7 − 4 = 3, so it starts with 3n.
    Adjustment: 4 − 3 = 1
    Leftover nth term = 3n + 1
  6. Put the two parts together. The nth term is the quadratic part plus the linear part.
    nth term = 2n2 + (3n + 1) = 2n2 + 3n + 1

Check. Test the formula. n = 2: 2 × 22 + 3 × 2 + 1 = 2 × 4 + 3 × 2 + 1 = 8 + 6 + 1 = 15 ✓   n = 5: 2 × 52 + 3 × 5 + 1 = 2 × 25 + 3 × 5 + 1 = 50 + 15 + 1 = 66 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.3
Find the nth term of the sequence 1, 10, 25, 46, 73, …

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

  1. Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
    Position n12345
    Term110254673
    First differences: 9, 15, 21, 27
    They are not all the same, so it is not a linear sequence.
  2. Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
    Terms1102546731st differences91521272nd differences666
    Second differences: 6, 6, 6
    They are all 6, so the sequence is quadratic.
  3. Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
    a = 6 ÷ 2 = 3
    The nth term starts with 3n2.
  4. Subtract 3n2 from each term. This removes the quadratic part. Work out 3n2 for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
    Position n12345
    n²1491625
    3n2312274875
    Term110254673
    Term − 3n2−2−2−2−2−2
  5. Look at what is left. The numbers left over are all the same, so there is no n term, just a number.
    Every remainder is −2, so b = 0 and c = −2.
  6. Put the two parts together. The nth term is the quadratic part plus the linear part.
    nth term = 3n2 − 2

Check. Test the formula. n = 2: 3 × 22 − 2 = 3 × 4 − 2 = 12 − 2 = 10 ✓   n = 5: 3 × 52 − 2 = 3 × 25 − 2 = 75 − 2 = 73 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.4
Find the nth term of the sequence 12, 19, 24, 27, 28, … (the terms go up and then level off).

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

  1. Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
    Position n12345
    Term1219242728
    First differences: 7, 5, 3, 1
    They are not all the same, so it is not a linear sequence.
  2. Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
    Terms12192427281st differences75312nd differences−2−2−2
    Second differences: −2, −2, −2
    They are all −2, so the sequence is quadratic.
  3. Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
    a = −2 ÷ 2 = −1
    The nth term starts with −n2.
  4. Subtract −n2 from each term. This removes the quadratic part. Work out −n2 for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
    Position n12345
    n²1491625
    −n2−1−4−9−16−25
    Term1219242728
    Term − −n21323334353
  5. Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
    Leftover sequence: 13, 23, 33, 43, 53
    Common difference = 23 − 13 = 10, so it starts with 10n.
    Adjustment: 13 − 10 = 3
    Leftover nth term = 10n + 3
  6. Put the two parts together. The nth term is the quadratic part plus the linear part.
    nth term = −n2 + (10n + 3) = −n2 + 10n + 3

Check. Test the formula. n = 2: −22 + 10 × 2 + 3 = −4 + 10 × 2 + 3 = −4 + 20 + 3 = 19 ✓   n = 5: −52 + 10 × 5 + 3 = −25 + 10 × 5 + 3 = −25 + 50 + 3 = 28 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.5
Find the nth term of the sequence 1, 3, 6, 10, 15, … (the triangular numbers).

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Find the second differences. Half of them gives the number in front of n². Subtract that part from the terms, and the leftover is linear.

  1. Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
    Position n12345
    Term1361015
    First differences: 2, 3, 4, 5
    They are not all the same, so it is not a linear sequence.
  2. Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
    Terms13610151st differences23452nd differences111
    Second differences: 1, 1, 1
    They are all 1, so the sequence is quadratic.
  3. Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
    a = 1 ÷ 2 = 12
    The nth term starts with 12n2.
  4. Subtract 12n2 from each term. This removes the quadratic part. Work out 12n2 for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
    Position n12345
    n²1491625
    12n2122928252
    Term1361015
    Term − 12n212132252
  5. Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
    Leftover sequence: 12, 1, 32, 2, 52
    Common difference = 1 − 12 = 12, so it starts with 12n.
    Adjustment: 12 − 12 = 0
    Leftover nth term = 12n
  6. Put the two parts together. The nth term is the quadratic part plus the linear part.
    nth term = 12n2 + (12n) = 12n2 + 12n

Check. Test the formula. n = 2: 12 × 22 + 12 × 2 = 12 × 4 + 12 × 2 = 2 + 1 = 3 ✓   n = 5: 12 × 52 + 12 × 5 = 12 × 25 + 12 × 5 = 252 + 52 = 15 ✓

Watch out. a is half of the second difference, not the second difference itself. And remember to subtract an² (using each position) before finding the linear part.

Worked example 11.8.6
The nth term of a sequence is n² + 2n. Is 80 a term of the sequence?

Key words. A quadratic sequence has an nth term with an n² in it, such as 2n² + 3. The first differences are the gaps between neighbouring terms. The second differences are the gaps between the first differences.

Method. Set the nth term equal to the number and solve for n. The number is in the sequence only if n is a positive whole number.

  1. Set the nth term equal to the number. The number is in the sequence only if some whole-number position n gives it, so we solve for n.
    n2 + 2n = 80
  2. Rearrange so that one side is 0. Subtract the number from both sides. A quadratic equation is easiest to solve when one side is zero.
    n2 + 2n − 80 = 0
  3. Factorise. Write it as two brackets multiplied together.
    Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -80.
    We need two numbers that multiply to -80 (the last number) and add to 2 (the middle number). List the pairs that multiply to -80:
    1 × -80 = -80,   add: 1 + -80 = -79 ✗
    -1 × 80 = -80,   add: -1 + 80 = 79 ✗
    2 × -40 = -80,   add: 2 + -40 = -38 ✗
    -2 × 40 = -80,   add: -2 + 40 = 38 ✗
    4 × -20 = -80,   add: 4 + -20 = -16 ✗
    -4 × 20 = -80,   add: -4 + 20 = 16 ✗
    5 × -16 = -80,   add: 5 + -16 = -11 ✗
    -5 × 16 = -80,   add: -5 + 16 = 11 ✗
    8 × -10 = -80,   add: 8 + -10 = -2 ✗
    -8 × 10 = -80,   add: -8 + 10 = 2 ✓ this pair works
    Put the pair into two brackets: (n − 8)(n + 10).
    Check by expanding: n² + 10n − 8n − 80 = n2 + 2n − 80. ✓
  4. Solve: one bracket must equal 0. If two things multiply to give 0 then at least one of them is 0.
    n − 8 = 0 gives n = 8
    n + 10 = 0 gives n = −10
  5. Decide. A position must be a whole number that is at least 1. Ignore a negative answer.
    Yes: 80 is the 8th term (the other answer is not a position).

Check. Substitute n = 8: 82 + 2 × 8 = 64 + 2 × 8 = 64 + 16 = 80. ✓

Watch out. A quadratic equation can have two answers. Keep only a positive whole number as the position.

Worked example 11.8.7
Here are the first four patterns of dots.
Pattern 1Pattern 2Pattern 3Pattern 4
Find the nth term, and the number of dots in pattern 20.

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. Count the dots, find the rule for the sequence of counts, and use the picture to see why it works.

  1. Count the dots in each pattern. Count carefully, one pattern at a time. This turns the pictures into a sequence of numbers.
    Pattern number n1234
    Number of dots381524
  2. Write the terms and find the first differences. The first differences are the gaps between neighbouring terms (later minus earlier). If they were all the same it would be an ordinary arithmetic sequence.
    Position n1234
    Term381524
    First differences: 5, 7, 9
    They are not all the same, so it is not a linear sequence.
  3. Find the second differences. Now find the gaps between the first differences. If these are all the same, the sequence is quadratic (it has an n² term).
    Terms3815241st differences5792nd differences22
    Second differences: 2, 2
    They are all 2, so the sequence is quadratic.
  4. Halve the second difference to find a, the number in front of n². The squares 1, 4, 9, 16 have first differences 3, 5, 7 and second differences 2. So n² has second difference 2. A sequence an² has second difference 2a. So a is half of the second difference.
    a = 2 ÷ 2 = 1
    The nth term starts with n2.
  5. Subtract n2 from each term. This removes the quadratic part. Work out n2 for each position, then take it away from the term. What is left must be a linear (arithmetic) sequence.
    Position n1234
    n²14916
    n214916
    Term381524
    Term − n22468
  6. Find the nth term of what is left. The leftover numbers form an arithmetic sequence. Find its common difference (that is the number in front of n), then the number to add, exactly as for a linear sequence.
    Leftover sequence: 2, 4, 6, 8
    Common difference = 4 − 2 = 2, so it starts with 2n.
    Adjustment: 2 − 2 = 0
    Leftover nth term = 2n
  7. Put the two parts together. The nth term is the quadratic part plus the linear part.
    nth term = n2 + (2n) = n2 + 2n
  8. Check that the rule makes sense with the picture. A good formula can be explained from the drawing.
    Pattern n is a rectangle of dots with n rows and n + 2 columns, so it has n(n + 2) = n² + 2n dots.
  9. Replace n by 20. Pattern 20 is the one in position 20.
    Pattern 20: 202 + 2 × 20 = 400 + 2 × 20 = 400 + 40 = 440

Check. Test the rule on patterns you can see. n = 1: 12 + 2 × 1 = 1 + 2 × 1 = 1 + 2 = 3 ✓   n = 4: 42 + 2 × 4 = 16 + 2 × 4 = 16 + 8 = 24 ✓

Watch out. Do not just keep adding on a drawing for a large pattern number. Find the nth term and substitute.

Watch out

a is half of the second difference, not the second difference itself. In 3, 8, 15, 24 the second difference is 2 so a = 1, not 2. Also remember to subtract an² using the position of each term: 1², 2², 3², … (that is 1, 4, 9, …).

Exercise 11H  Quadratic sequences

Non-calculator
1
FFind the nth term of each quadratic sequence.
(a)3, 8, 15, 24, 35, …
(b)5, 8, 13, 20, 29, …
(c)2, 8, 16, 26, 38, …
(d)5, 7, 11, 17, 25, …
(e)6, 13, 22, 33, 46, …
(f)0, 3, 8, 15, 24, …
2
FFind the nth term of each quadratic sequence.
(a)6, 15, 28, 45, 66, …
(b)6, 14, 28, 48, 74, …
(c)1, 13, 33, 61, 97, …
(d)7, 18, 33, 52, 75, …
(e)6, 23, 50, 87, 134, …
(f)4, 13, 28, 49, 76, …
3
FFind the nth term of each sequence. The second difference is negative, or a is a fraction.
(a)12, 19, 24, 27, 28, …
(b)18, 12, 2, −12, −30, …
(c)1, 3, 6, 10, 15, …
(d)3, 6, 10, 15, 21, …
(e)13, 18, 17, 10, −3, …
4
FHere are the first four patterns of dots.
Pattern 1Pattern 2Pattern 3Pattern 4
(a)Find the nth term for the number of dots.
(b)How many dots are in pattern 15?
5
FHere are the first four patterns of dots.
Pattern 1Pattern 2Pattern 3Pattern 4
(a)Find the nth term.
(b)Which pattern has exactly 80 dots?
6
PSA square pond (blue) is surrounded by a path (orange). The pictures show the first three patterns. The total number of tiles, blue and orange together, is counted.
Pattern 1Pattern 2Pattern 3
(a)Find the nth term for the total number of tiles.
(b)How many tiles are there in total in pattern 12?
7
FFor each quadratic sequence, find the nth term and then the term asked for.
(a)4, 10, 20, 34, 52, …   Find the 10th term.
(b)2, 8, 18, 32, 50, …   Find the 12th term.
(c)0, 3, 8, 15, 24, …   Find the 20th term.
8
FIs each number a term of the sequence with the given nth term? If so, say which term.
(a)n2 + 3: is 52 a term?
(b)n2 + 3: is 50 a term?
(c)n2 + 2n: is 80 a term?
(d)2n2 + 1: is 99 a term?
(e)n2 + 2n: is 79 a term?
(f)3n2 − 2: is 73 a term?
9
RSam says the nth term of 2, 6, 12, 20, … is 4n − 2. Show that Sam is wrong and find the correct nth term.
10
RShow that the nth term of the sequence 2, 5, 10, 17, 26, … is n² + 1.
11
RThe nth term of a sequence is n² + n. Show that every term of the sequence is an even number.
12
RShow that the difference between the (n + 1)th square number and the nth square number is 2n + 1.
13
PSThe nth term of a sequence is n² − 4n + 7. Work out the first five terms. Which is the smallest term?
14
RThe first four terms of a quadratic sequence are 5, 12, 23, 38.
(a)Find the nth term.
(b)Is 500 a term of this sequence? Show how you decide.
11.9

Sequences in context, reasoning and proof

Edexcel 1MA1: A23, A24, A25
Grade 4-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Sequences in context, reasoning and proofComing soon

In the exam, sequences often come wrapped in a story: a taxi fare, a saving plan, rows of seats, tiles in a patio. A good plan makes these problems easy.

  1. Write the first few terms. Turn the words or picture into a list of numbers.
  2. Decide what kind of sequence it is. Check the differences (arithmetic), then the ratios (geometric), then the second differences (quadratic, Higher).
  3. Find the rule or the nth term. Use the common difference method from Section 11.4.
  4. Answer the question. Substitute for n to find a term, or set the nth term equal to a number and solve to find a position.
  5. Check by testing your rule on terms you already know, and make sure the answer makes sense in the context (you cannot have 2.5 chairs).

Explaining and proving. The nth term lets you prove things about every term, not just the first few. To show that every term of 6n + 3 is a multiple of 3, write it as 3(2n + 1): a 3 times a whole number. To show that a number cannot be in a sequence, show that it does not fit the pattern of the terms. Checking a few values is not a proof; the algebra must work for every n.

Sequences also let us compare two situations. If A costs 8n + 20 and B costs 11n + 5, set the two nth terms equal to find when they cost the same, and use the sequences to say which is cheaper before and after that point.

Pattern 1Pattern 2Pattern 3Pattern 4
Each pattern is a square. The orange dots are the L-shape added to the previous square: 1, 3, 5, 7, … (the odd numbers).

Key ideas

  • Plan: list terms, decide the type, find the rule or nth term, answer, then check.
  • Substitute to find a term; set the nth term equal to a number and solve to find a position.
  • A proof must work for every n: use algebra, such as 6n + 3 = 3(2n + 1).
  • Give the answer in the context of the question, with units, and check it makes sense.
Worked example 11.9.1
A plumber charges a £40 call-out fee plus £25 for each hour of work. The cost of a job lasting 1, 2, 3, 4 hours is £65, £90, £115, £140. Find the nth term and the cost of an 8-hour job.

Key words. An arithmetic sequence goes up or down by the same amount every time. That amount is the common difference, written d. The nth term is a formula that gives any term from its position n.

Method. Use the common difference to start the nth term (it is the number in front of n), then work out what has to be added or subtracted.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term6590115140
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    90 − 65 = 25
    115 − 90 = 25
    140 − 115 = 25
    Common difference d = 25
  3. Write the 25 times table underneath. Each time the position goes up by 1, the term changes by 25. That is just like the 25 times table: 25, 50, 75, … So the nth term must start with 25n.
    Position n1234
    Term6590115140
    25n255075100
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term6590115140
    25n255075100
    Term − 25n40404040
    The adjustment is always 40. So we must add 40 to 25n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 25n + 40
    Shortcut for later: the adjustment 40 is the term before the first term, 65 − 25 = 40.
  6. Replace n by 8 to find the 8th term. The 8th term is the one in position 8.
    8th term = 25 × 8 + 40 = 200 + 40 = 240

Check. Test the formula on terms we already know. n = 1: 25 × 1 + 40 = 25 + 40 = 65 ✓   n = 4: 25 × 4 + 40 = 100 + 40 = 140 ✓

Watch out. The number in front of n is the common difference, not the first term. Check with the first term: if your formula does not give it, something is wrong.

Worked example 11.9.2
Using the same costs (£65, £90, £115, £140, …), could a job cost exactly £290? If so, how many hours does it last?

Key words. The position of a term is its place in the sequence: 1st, 2nd, 3rd and so on. We use the letter n for the position number. The nth term (or position-to-term rule) is a formula that gives the term for any position n.

Method. A number is in a sequence only if it is the term in a whole-number position. Set the nth term equal to the number and solve for n.

  1. Write the position numbers above the terms. The position is the place of each term in the list: 1st, 2nd, 3rd, and so on. The letter n stands for any position.
    Position n1234
    Term6590115140
  2. Find the common difference d. Subtract each term from the next one. In an arithmetic sequence you always get the same answer, and that answer is called the common difference.
    90 − 65 = 25
    115 − 90 = 25
    140 − 115 = 25
    Common difference d = 25
  3. Write the 25 times table underneath. Each time the position goes up by 1, the term changes by 25. That is just like the 25 times table: 25, 50, 75, … So the nth term must start with 25n.
    Position n1234
    Term6590115140
    25n255075100
  4. Compare the terms with the times table. The times table has the right steps but may start in the wrong place. Work out (term − times table) for each position. It should be the same number every time.
    Term6590115140
    25n255075100
    Term − 25n40404040
    The adjustment is always 40. So we must add 40 to 25n.
  5. Put it together and write the nth term. The nth term is the times table part plus the adjustment.
    nth term = 25n + 40
    Shortcut for later: the adjustment 40 is the term before the first term, 65 − 25 = 40.
  6. Understand what the question asks. A number is in the sequence only if some position n gives it. Positions are whole numbers: 1, 2, 3, … So we need to find out whether n comes out as a whole number.
    nth term = 25n + 40
    Is there a whole number n with 25n + 40 = 290?
  7. Set the nth term equal to the number and solve for n. Subtract 40 on both sides to get the n term on its own, then divide by 25.
    25n + 40 = 290
    25n = 290 − 40 = 250
  8. Divide by 25. The inverse of multiplying by 25 is dividing by 25.
    n = 250 ÷ 25 = 10
  9. Decide. 10 is a whole number and it is at least 1, so it is a real position.
    Yes: 290 is the 10th term.

Check. Substitute n = 10: 25 × 10 + 40 = 250 + 40 = 290, which is 290. ✓

Watch out. Do not just say "it goes up in 4s so it is in the sequence". A number can be close to a term and still not be one. Work out n and check that it is a whole number.

Worked example 11.9.3
Sequence A is 4, 7, 10, 13, … (nth term 3n + 1). Sequence B is 30, 28, 26, 24, … (nth term 32 − 2n). In which position is a term of A first bigger than the term of B?

Method. Make an inequality: A's term > B's term. Solve it, then round up to a whole number.

  1. Write the inequality. We want the term of A to be bigger than the term of B in the same position n.
    3n + 1 > 32 − 2n
  2. Add 2n to both sides. This gets all the n terms on the left, keeping the inequality balanced.
    5n + 1 > 32
  3. Subtract 1 from both sides.
    5n > 31
  4. Divide by 5.
    n > 31 ÷ 5 = 6.2
  5. Choose the first whole number. The next whole number above 6.2 is 7.
    n = 7
  6. Test positions 6 and 7. The 6th terms and 7th terms:
    n = 6: A = 3 × 6 + 1 = 19, B = 32 − 12 = 20 (A is smaller)
    n = 7: A = 3 × 7 + 1 = 22, B = 32 − 14 = 18 (A is bigger)

Check. In position 7, A = 22 and B = 18, so A is bigger, and in position 6 it is not. ✓

Watch out. Round n up to the next whole number, and test the position before it.

Worked example 11.9.4
The nth term of a sequence is 6n + 3. Show that every term is an odd multiple of 3.

Method. Write 6n + 3 with a factor of 3 outside a bracket, then think about even and odd.

  1. Factorise. Both 6n and 3 divide by 3.
    6n + 3 = 3(2n + 1)
  2. Look at 3 × (a whole number). 2n + 1 is a whole number, so every term is 3 times a whole number: a multiple of 3.
    3(2n + 1) is a multiple of 3.
  3. Show it is odd. 2n is even, so 2n + 1 is odd. An odd number times 3 is odd.
    Every term is an odd multiple of 3.

Check. The terms are 9, 15, 21, 27: all odd multiples of 3. ✓

Watch out. Examples are not proof. The factorised form works for every n.

Worked example 11.9.5
The nth term of a sequence is 2n + 1. Show that the sum of any two consecutive terms is a multiple of 4.

Method. Write the two consecutive terms using n and n + 1, add them and factorise.

  1. Write the two terms. The next term after the nth term uses n + 1 in place of n.
    nth term = 2n + 1
    (n + 1)th term = 2(n + 1) + 1 = 2n + 3
  2. Add them. Collect like terms.
    (2n + 1) + (2n + 3) = 4n + 4
  3. Factorise. Both terms divide by 4.
    4n + 4 = 4(n + 1)
  4. Conclude. n + 1 is a whole number, so the sum is 4 times a whole number.
    The sum of any two consecutive terms is a multiple of 4.

Check. n = 3: 7 + 9 = 16 = 4 × 4. ✓

Watch out. Use n and n + 1 for consecutive terms.

Watch out

Always put your answer back into the story. If the nth term gives n = 24.5 hours or 6.3 chairs, the answer to “could it happen?” is no. And a proof needs a general argument: three or four correct examples do not prove something is true for every term.

Exercise 11I  Sequences in context and reasoning

Non-calculator
1
PSRhys earns £2150 a month in year 1 at a company, £2300 a month in year 2, £2450 a month in year 3, and so on (a £150 rise each year). Find the nth term and his monthly pay in year 10.
2
PSA tank holds 500 litres. Water drains out at 20 litres every minute. Term 1 is 500 (the start), term 2 is 480 (after 1 minute), term 3 is 460, and so on. After how many minutes is the tank empty?
3
PSSequence A is 4, 7, 10, 13, … and sequence B is 30, 28, 26, 24, … In which position does A first have a bigger term than B? Show your working.
4
PSA stack of identical chairs is 90 cm high with one chair. Each extra chair adds 8 cm, so a stack of n chairs is (8n + 82) cm high. The roof of a van is 300 cm high. What is the height of the tallest stack that will fit, and how many chairs is that?
5
PSA theatre has 14 seats in the first row, 16 in the second row, 18 in the third row, and so on. The number of seats in row n is 2n + 12.
(a)How many seats are in row 20?
(b)A row has 40 seats. Which row is it?
6
PSA video gets 5 views on day 1, 15 on day 2 and 45 on day 3. The number of views triples each day. How many views does it get on day 5?
7
RShow that every term of the sequence with nth term 6n + 3 is a multiple of 3.
8
RMia says the nth term of 10, 7, 4, 1, … is 3n + 7. What mistake has she made? Find the correct nth term.
9
REach pattern in the picture is a square of dots. The orange dots show the new dots added to make each square. Explain why the number of orange dots in pattern n is 2n − 1, and use this to explain why 1 + 3 + 5 + 7 = 4².
Pattern 1Pattern 2Pattern 3Pattern 4
10
FFill in the table for each description and say whether the sequence is arithmetic, geometric or quadratic.
(a)The nth term is 5n + 1
(b)The nth term is 3 × 2n
(c)The nth term is n² + 1
11
PSThe 4th term of an arithmetic sequence is 17 and the 10th term is 41. Find the nth term and the first term.
12
PSHire company A charges £20 plus £8 per day, so n days cost 8n + 20 pounds. Company B charges £5 plus £11 per day, so n days cost 11n + 5. After how many days do the two companies charge the same? Which is cheaper for 10 days?
13
RThe nth term of a sequence is 2n. Show that each term is double the previous term.

Chapter summary

SkillMethod
Term-to-term ruleFind the differences (add/subtract), then the ratios (multiply/divide); apply the rule to the last term
Position-to-term ruleReplace n by the position and follow BIDMAS; the first term is n = 1
Pattern sequencesCount each pattern, find the rule, explain it with the picture, continue the sequence
nth term of an arithmetic sequenceFind d, write the d times table, subtract to find the adjustment: dn + (first term − d)
Is a number in the sequence?Solve nth term = number; it is a term only if n is a positive whole number
First term above (below) a valueSolve an inequality; round n up (down) to a whole number and test the term before
Special sequencesSquares n², cubes n³, triangular n(n + 1) ÷ 2, powers of 2: 2n, Fibonacci-type: add the previous two
Geometric sequencesCommon ratio r = term ÷ previous term; powers rn; Higher: nth term arn − 1; surds: √k × √k = k
Two terms given (Higher)d = (change in terms) ÷ (change in positions), then find the adjustment
Quadratic sequences (Higher)Second differences constant; a = half the second difference; subtract an²; find the linear part bn + c
Proving with sequencesUse the nth term and algebra, e.g. 6n + 3 = 3(2n + 1)

Can you…?

  • I can continue a sequence and describe its term-to-term rule in words.
  • I can find terms from a position-to-term rule, including powers such as n² and 2n.
  • I can continue patterns made from matchsticks, tiles and dots.
  • I can find the nth term of an arithmetic sequence.
  • I can decide whether a number is in a sequence and find the first term above or below a value.
  • I can recognise square, cube, triangular, Fibonacci-type and geometric sequences.
  • I can find the common ratio of a geometric sequence and continue it.
  • Higher: I can find the nth term of a quadratic sequence using second differences.
  • Higher: I can find the nth term of a geometric sequence and work with a surd ratio.
  • I can use the nth term to explain or prove something about every term.

Key terms

Sequence
A list of numbers that follow a rule.
Term
One number in a sequence.
Term-to-term rule
A rule that tells you how to get from one term to the next.
Position-to-term rule (nth term)
A formula that gives the term from its position n.
Position
The place of a term in the sequence: 1st, 2nd, 3rd, …
Arithmetic (linear) sequence
A sequence that goes up or down by the same amount each time.
Common difference
The amount added each time in an arithmetic sequence.
Geometric sequence
A sequence made by multiplying by the same number each time.
Common ratio
The number multiplied each time in a geometric sequence.
Square number
A number multiplied by itself: 1, 4, 9, 16, …
Cube number
A number multiplied by itself twice more: 1, 8, 27, 64, …
Triangular number
The numbers 1, 3, 6, 10, 15, … found by adding 1, then 2, then 3, …
Fibonacci-type sequence
A sequence where each term is the sum of the two terms before it.
First differences
The differences between neighbouring terms.
Second differences
The differences between the first differences.
Quadratic sequence
A sequence whose nth term contains n²; its second differences are constant (Higher).
Surd
A root that is not exact, such as √2 (Higher).

Exercise 11Z  Check-up (non-calculator)

Non-calculator
1
Write down the next two terms of 4, 11, 18, 25, …
2
Describe the term-to-term rule of the sequence 5, 15, 45, 135, …
3
The nth term of a sequence is 5n − 3. Write down the first four terms.
4
Find the nth term of the sequence 7, 11, 15, 19, …
5
Find the nth term of the sequence 30, 27, 24, 21, …
6
The nth term of a sequence is 3n + 1. Is 100 a term of the sequence?
7
Work out the 12th triangular number.
8
Here is a Fibonacci-type sequence: 2, 3,  ? , 8,  ?  Write down the two missing numbers.
9
The nth term of a sequence is 7n − 4. Find the first term that is greater than 100.
10
Here are the first three patterns of triangles made from matchsticks. Find the nth term for the number of matchsticks.
Pattern 1Pattern 2Pattern 3

Total 43 marks. Allow about 43 minutes.

Exercise 11X  Exam-style questions

Non-calculator
1
FWrite down the next two terms of the sequence 8, 13, 18, 23, …[2 marks]
2
FWrite down the term-to-term rule for the sequence 2, 6, 18, 54, …[1 mark]
3
FThe nth term of a sequence is 4n + 3. Write down the first three terms.[2 marks]
4
FFind the nth term of the sequence 6, 11, 16, 21, …[2 marks]
5
RThe nth term of a sequence is 4n − 1. Is 150 a term of this sequence? You must show your working.[3 marks]
6
PSHere are the first three patterns made from matchsticks. Find the nth term, and the number of matchsticks in pattern 40.[3 marks]
Pattern 1Pattern 2Pattern 3
7
FHere is a Fibonacci-type sequence: 3, 4,  ? , 11,  ?  Find the two missing numbers.[2 marks]
8
PSThe nth term of a sequence is 7n − 4. Find the first term of this sequence that is greater than 200.[3 marks]
9
FWork out the 8th square number and the 5th cube number.[2 marks]
(a)The 8th square number
(b)The 5th cube number
10
PSA van hire company charges a fixed fee plus a daily rate. The cost, in pounds, for 1, 2, 3, 4 days is 42, 54, 66, 78. Find the nth term of this sequence and use it to work out the cost of hiring the van for 15 days.[4 marks]
11
FThe first term of a sequence is 4. The rule is “double the term and then subtract 1”. Write down the next two terms.[2 marks]
12
PSHere are the first three tile patterns.[4 marks]
Pattern 1Pattern 2Pattern 3
(a)Find the nth term for the number of tiles.
(b)Which pattern has 101 tiles?
13
FFind the nth term of the quadratic sequence 2, 7, 14, 23, 34, …[3 marks]
14
PSIn a geometric sequence the 2nd term is 12 and the 5th term is 96. All the terms are positive. Find the common ratio and the nth term of the sequence.[4 marks]
15
FThe first two terms of a geometric sequence are 3 and 3√2. Write down the next two terms. Give each term as simply as possible.[3 marks]
16
RThe nth term of a sequence is 2n + 1. Prove that the sum of any two consecutive terms is a multiple of 4.[3 marks]
← Chapter 10: InequalitiesChapter 12: Straight-line graphs and coordinates →

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