GCSE Maths: Sequences in context, reasoning and proof (Higher)
Part of the GCSE Maths sequences and nth term chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Sequences in context, reasoning and proof
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.
- Write the first few terms. Turn the words or picture into a list of numbers.
- Decide what kind of sequence it is. Check the differences (arithmetic), then the ratios (geometric), then the second differences (quadratic, Higher).
- Find the rule or the nth term. Use the common difference method from Section 11.4.
- 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.
- 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.
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.
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.
- 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 n 1 2 3 4 Term 65 90 115 140 - 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 - 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 n 1 2 3 4 Term 65 90 115 140 25n 25 50 75 100 - 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.
The adjustment is always 40. So we must add 40 to 25n.Term 65 90 115 140 25n 25 50 75 100 Term − 25n 40 40 40 40 - 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. - 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.
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.
- 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 n 1 2 3 4 Term 65 90 115 140 - 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 - 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 n 1 2 3 4 Term 65 90 115 140 25n 25 50 75 100 - 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.
The adjustment is always 40. So we must add 40 to 25n.Term 65 90 115 140 25n 25 50 75 100 Term − 25n 40 40 40 40 - 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. - 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? - 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 - Divide by 25. The inverse of multiplying by 25 is dividing by 25.n = 250 ÷ 25 = 10
- 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.
Method. Make an inequality: A's term > B's term. Solve it, then round up to a whole number.
- 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
- Add 2n to both sides. This gets all the n terms on the left, keeping the inequality balanced.5n + 1 > 32
- Subtract 1 from both sides. 5n > 31
- Divide by 5. n > 31 ÷ 5 = 6.2
- Choose the first whole number. The next whole number above 6.2 is 7.n = 7
- 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.
Method. Write 6n + 3 with a factor of 3 outside a bracket, then think about even and odd.
- Factorise. Both 6n and 3 divide by 3.6n + 3 = 3(2n + 1)
- 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.
- 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.
Method. Write the two consecutive terms using n and n + 1, add them and factorise.
- 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 - Add them. Collect like terms.(2n + 1) + (2n + 3) = 4n + 4
- Factorise. Both terms divide by 4.4n + 4 = 4(n + 1)
- 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-calculatorKey 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.
- 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 n 1 2 3 4 Term 2150 2300 2450 2600 - 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.2300 − 2150 = 150
2450 − 2300 = 150
2600 − 2450 = 150
Common difference d = 150 - Write the 150 times table underneath. Each time the position goes up by 1, the term changes by 150. That is just like the 150 times table: 150, 300, 450, … So the nth term must start with 150n.
Position n 1 2 3 4 Term 2150 2300 2450 2600 150n 150 300 450 600 - 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.
The adjustment is always 2000. So we must add 2000 to 150n.Term 2150 2300 2450 2600 150n 150 300 450 600 Term − 150n 2000 2000 2000 2000 - Put it together and write the nth term. The nth term is the times table part plus the adjustment.nth term = 150n + 2000
Shortcut for later: the adjustment 2000 is the term before the first term, 2150 − 150 = 2000. - Replace n by 10 to find the 10th term. The 10th term is the one in position 10.10th term = 150 × 10 + 2000 = 1500 + 2000 = 3500
Check. Test the formula on terms we already know. n = 1: 150 × 1 + 2000 = 150 + 2000 = 2150 ✓ n = 4: 150 × 4 + 2000 = 600 + 2000 = 2600 ✓
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.
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.
- 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 n 1 2 3 4 Term 500 480 460 440 - 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.480 − 500 = −20
460 − 480 = −20
440 − 460 = −20
Common difference d = −20 - Write the −20 times table underneath. Each time the position goes up by 1, the term changes by −20. That is just like the −20 times table: −20, −40, −60, … So the nth term must start with −20n.
Position n 1 2 3 4 Term 500 480 460 440 −20n −20 −40 −60 −80 - 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.
The adjustment is always 520. So we must add 520 to −20n.Term 500 480 460 440 −20n −20 −40 −60 −80 Term − −20n 520 520 520 520 - Put it together and write the nth term. The nth term is the times table part plus the adjustment.nth term = −20n + 520 (you can also write it as 520 − 20n)
Shortcut for later: the adjustment 520 is the term before the first term, 500 − (−20) = 520. - 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 = 520 − 20n
Is there a whole number n with 520 − 20n = 0? - Set the nth term equal to the number and solve for n. Subtract 520 on both sides to get the n term on its own, then divide by −20.520 − 20n = 0
−20n = 0 − 520 = −520 - Divide by (−20). The inverse of multiplying by −20 is dividing by −20.n = −520 ÷ (−20) = 26
- Decide. 26 is a whole number and it is at least 1, so it is a real position.Yes: 0 is the 26th term.
Check. Substitute n = 26: 520 − 20 × 26 = 520 − 520 = 0, which is 0. ✓
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.
Method. Find the nth term of each sequence, then compare terms in the same position.
- Find the nth term of A. Common difference 3, adjustment 4 − 3 = 1.A: 3n + 1
- Find the nth term of B. Common difference −2, adjustment 30 + 2 = 32.B: 32 − 2n
- Write A > B and solve. Add 2n to both sides, subtract 1, divide by 5.3n + 1 > 32 − 2n
5n > 31
n > 6.2 - Choose the whole number. The next whole number above 6.2.n = 7
- Check. Work out both terms in positions 6 and 7.Position 6: A = 19, B = 20 (A is smaller)
Position 7: A = 22, B = 18 (A is bigger)
Check. Listing the terms: A: 4, 7, 10, 13, 16, 19, 22; B: 30, 28, 26, 24, 22, 20, 18. A first exceeds B at position 7. ✓
Watch out. Test the position before as well, to show it is the first.
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.
- Write what we want as an inequality. We want terms that are less than 300.8n + 82 < 300
- Solve it like an equation. Subtract 82 on both sides, then divide both sides by 8.8n < 218
n < 218 ÷ 8 = 27.25 - Choose the largest whole number that works. n must be a whole number less than 27.25.n = 27
- Work out the term and test the next one. The term in position n must be less than 300, and the next one must not be.27th term = 8 × 27 + 82 = 216 + 82 = 298
28th term = 8 × 28 + 82 = 224 + 82 = 306 (too big)
Check. 298 < 300 but 306 > 300 or equal. ✓
Watch out. Round down here, because the term must stay below the limit.
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.
- Write the rule. This formula gives the term for any position n.nth term = 2n + 12
- The position we want is 20, so replace n by 20. Every n in the formula becomes the same number.20th term = 2 × 20 + 12
- Work it out, doing powers first, then multiplying, then adding or subtracting. This is the BIDMAS order.20th term = 2 × 20 + 12 = 40 + 12 = 52
- Write the answer. The 20th term is 52.
Check. The 20th term should be the first term plus 19 steps of 2: 14 + 19 × 2 = 52. ✓
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.
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.
- 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 = 2n + 12
Is there a whole number n with 2n + 12 = 40? - Set the nth term equal to the number and solve for n. Subtract 12 on both sides to get the n term on its own, then divide by 2.2n + 12 = 40
2n = 40 − 12 = 28 - Divide by 2. The inverse of multiplying by 2 is dividing by 2.n = 28 ÷ 2 = 14
- Decide. 14 is a whole number and it is at least 1, so it is a real position.Yes: 40 is the 14th term.
Check. Substitute n = 14: 2 × 14 + 12 = 28 + 12 = 40, which is 40. ✓
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.
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 5th term.
- Try the differences first. We always look for "add or subtract the same number" before anything else.15 − 5 = 10
45 − 15 = 30
The differences are not all the same, so it is not an add/subtract rule. - 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.15 ÷ 5 = 3
45 ÷ 15 = 3 - Look at the answers. They are all the same, so we multiply by that number each time.The rule is multiply by 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.45 × 3 = 135
135 × 3 = 405 - Pick out the term you want. The 5th term is the one in position 5. Count along the list.5, 15, 45, 135, 405
The 5th term is 405.
Check. Look back along the whole list: 15, 45, 135, 405. 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.
Method. Factorise the nth term.
- Find the common factor. 6 and 3 both divide by 3.6n + 3 = 3 × 2n + 3 × 1
- Write it with a bracket. Take the 3 outside.6n + 3 = 3(2n + 1)
- Conclude. 2n + 1 is a whole number, so 3 times it is a multiple of 3.Every term is a multiple of 3.
Check. n = 5: 6 × 5 + 3 = 33 = 3 × 11. ✓
Watch out. Show the factor of 3 in the algebra; do not just try some values.
Method. Test Mia's rule and then find the correct one.
- Test Mia's rule. Replace n by 1.3 × 1 + 7 = 10 ✓ but 3 × 2 + 7 = 13, and the 2nd term is 7 ✗
- Find the common difference. Subtract each term from the next.7 − 10 = −3, 4 − 7 = −3: d = −3 (the sequence goes down)
- Find the adjustment. First term minus d.10 − (−3) = 13
- Write the nth term. −3n + 13 (or 13 − 3n)
Check. n = 1: −3 + 13 = 10 ✓; n = 4: −12 + 13 = 1 ✓
Watch out. A sequence that goes down has a negative number in front of n.
Method. Count the dots in the L-shape in terms of n, then see how the L-shapes fit together.
- Count one L-shape. Pattern n is a square with n dots on each side. The L has one full side and one full bottom row.n dots down the side + n dots along the bottom
- Allow for the corner. The corner dot is in both the side and the bottom, so it has been counted twice.Number of dots = n + n − 1 = 2n − 1
- Check with the pictures. Put n = 1, 2, 3, 4 into 2n − 1.1, 3, 5, 7 ✓
- Build the square. Pattern 4 is made of all four L-shapes together: pattern 1 (1 dot) + 3 + 5 + 7.1 + 3 + 5 + 7 = 16 = 4 × 4 = 4²
Check. Pattern 3 has 3 × 3 = 9 dots and 1 + 3 + 5 = 9. ✓
Watch out. The sum of the first n odd numbers is always n².
Method. Work out the first few terms, then look at the differences and ratios.
- Find the first terms. Replace n by 1, 2, 3, 4.6, 11, 16, 21
- Test the differences. Subtract each term from the next.11 − 6 = 5, 16 − 11 = 5, 21 − 16 = 5: all the same, so it is arithmetic.
Check. The number in front of n is 5. ✓
Watch out. Always test both the differences and the ratios.
Method. Work out the first few terms, then look at the differences and ratios.
- Find the first terms. Work out 2n first, then multiply by 3.3 × 2 = 6, 3 × 4 = 12, 3 × 8 = 24, 3 × 16 = 48
- Test the ratios. Divide each term by the one before.12 ÷ 6 = 2, 24 ÷ 12 = 2, 48 ÷ 24 = 2: all the same, so it is geometric.
Check. Each term is double the last. ✓
Watch out. Always test both the differences and the ratios.
Method. Work out the first few terms, then look at the differences and ratios.
- Find the first terms. Square first, then add 1.1 + 1 = 2, 4 + 1 = 5, 9 + 1 = 10, 16 + 1 = 17
- Test the differences. Subtract each term from the next.3, 5, 7: not the same. The differences go up by 2, so the second differences are constant: quadratic.
Check. Second differences: 2, 2. ✓
Watch out. Always test both the differences and the ratios.
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.
- Find how many steps there are between the two terms. Going from the 4th term to the 10th term means taking steps of the common difference d. The number of steps is the difference between the positions.10 − 4 = 6 steps
- Find the total change in the terms. This is how much the sequence has gone up (or down) over those steps.41 − 17 = 24
- Divide to find the common difference. 6 equal steps make a total change of 24, so one step is the total divided by 6.d = 24 ÷ 6 = 4
- Use the common difference to start the nth term. The nth term of an arithmetic sequence starts with dn.nth term = 4n + (something)
- Put the 4th term into the formula to find the something. We know that when n = 4 the term is 17. Let the missing number be c.4n + c with n = 4: 4 × 4 + c = 17
16 + c = 17
c = 17 − 16 = 1 - Write the nth term. nth term = 4n + 1
Check. Test both terms. n = 4: 4 × 4 + 1 = 16 + 1 = 17 ✓ n = 10: 4 × 10 + 1 = 40 + 1 = 41 ✓
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.
Method. Make the two nth terms equal to find when the costs match, then compare at 10 days.
- Make the costs equal. Set the two nth terms equal to each other and solve for n.8n + 20 = 11n + 5
20 − 5 = 11n − 8n
15 = 3n
n = 5 - Find the cost at 5 days. Put n = 5 into either rule.8 × 5 + 20 = 60 and 11 × 5 + 5 = 60: both £60.
- Compare at 10 days. Work out each cost for n = 10.A: 8 × 10 + 20 = £100
B: 11 × 10 + 5 = £115 - Answer. A is the cheaper by £15.A is cheaper for 10 days.
Check. B has the smaller fixed fee, so it is cheaper for fewer days, but its daily rate is higher, so A wins after day 5. ✓
Watch out. Give the answer in words in the context.
Method. Write the (n + 1)th term and compare it with the nth term.
- Write the two terms. Replace n by n + 1 for the next term.nth term = 2n
(n + 1)th term = 2n + 1 - Use the laws of indices. 2n + 1 means n + 1 twos multiplied: one more 2 than 2n.2n + 1 = 2n × 21 = 2 × 2n
- Conclude. So each term is double the term before it.
Check. n = 3: 24 = 16 = 2 × 8. ✓
Watch out. 2n + 1 is not 2n + 1.