Printable · GCSE Foundation · ages 14-16
Generating sequences worksheet — GCSE Foundation
Fifteen questions on "generating sequences" — DfE statement A23. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Generating sequences worksheet — GCSE Foundation
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- 1.The nth term of a sequence is 2n² + 1. Work out the 4th term of the sequence.
- 2.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 3.A sequence has the position-to-term rule: the nth term is 3n. Write down the first four terms of the sequence.
- 4.A sequence begins at 7. Each term after the first is found by adding 6 to the term before it. Work out the 6th term of the sequence.
- 5.A sequence has the position-to-term rule: the term is the position number multiplied by itself. Its first five terms are 1, 4, 9, 16, 25. Work out the 8th term.
- 6.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 7.A sequence begins at 50, and each term after that is found by subtracting 8 from the term before it. Priya says the 8th term of the sequence is negative. Is Priya correct?
- 8.In a sequence, each term is half the term before it. The sequence begins 100, 50, 25. Write down the next term.
- 9.A wall is tiled in rows, and every row uses the same number of tiles. One row uses 10 tiles, two rows use 20 tiles and three rows use 30 tiles. Work out how many tiles are needed for 7 rows.
- 10.A sequence has the position-to-term rule n² + 1, where n is the position number. Work out how much bigger the 5th term is than the 4th term.
- 11.The nth term of a sequence is 3n + 2. Work out the 4th term of the sequence.
- 12.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 13.A sequence has the position-to-term rule n² − 3, where n is the position number. Work out the difference between the 6th term and the 5th term.
- 14.Here are the first four terms of an arithmetic sequence: 5, 9, 13, 17. Work out the 5th term.
- 15.A sequence has the position-to-term rule: the term is the square of the position number. Its first four terms are 1, 4, 9, 16. Work out the 10th term.
Answer key
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
- (a) 11 — Substitute n=6 into 2n−1: 2×6−1=11. A candidate who adds 2 and 6 and then subtracts 1, instead of multiplying 2 by 6 first, would compute 2+6−1=7. A candidate who substitutes the wrong term number, n=5, would reach 2×5−1=9. A candidate who forgets to subtract 1 would compute just 2×6=12.
- (b) 3, 6, 9, 12 — Method: substitute the positions n = 1, 2, 3 and 4 into the rule in turn, because a position-to-term rule gives each term from its own position number. Working: 3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9 and 3 × 4 = 12. Answer: 3, 6, 9, 12. The distractors: 3, 9, 27, 81 comes from reading 3n as 3 multiplied by itself n times and so multiplying by 3 at every step; 0, 3, 6, 9 comes from starting the count at n = 0, which shifts every term one place; 4, 5, 6, 7 comes from reading 3n as n + 3 and adding 3 to each position number instead of multiplying by 3.
- (b) 37 — The terms are 7, 13, 19, 25, 31, 37 — each found by adding 6 to the term before, so the 6th term is 37. Adding 6 six times to the first term instead of five times, 7 + 6 × 6 = 43, treats the first term as if it were before the sequence starts. Stopping one term early gives the 5th term, 31. Stopping two terms early gives the 4th term, 25.
- (b) 64 — Method: a position-to-term rule is applied straight to the position number, so a term far along the sequence can be found without writing out the terms in between. Working: the rule multiplies the position by itself and the position asked for is 8, so the term is 8 × 8. Answer: 64. The distractors: 16 comes from multiplying the position by 2 instead of by itself; 81 comes from working out 9 × 9, one position too far along; 8 comes from writing down the position number itself as the term.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (d) Yes: the 8th term is 50 − 8 × 7 = −6, which is negative. — Method: find the 8th term by subtracting 8 a total of 7 times from the first term, since the 1st term itself needs 0 subtractions. Working: 8th term = 50 − 8 × 7 = 50 − 56 = −6, which is negative, so Priya is correct. Answer: Yes, the 8th term is 50 − 8 × 7 = −6, which is negative. The "50 − 8 × 6 = 2" option subtracts 8 only six times instead of seven, an off-by-one error in counting the steps. The "50 − 8 × 8 = −14" option subtracts 8 eight times instead of seven, the opposite off-by-one error. The claim that repeated subtraction "can never go negative" ignores that subtracting enough times from any starting value eventually gives a negative result.
- (b) 12.5 — Method: apply the term-to-term rule to the term just written, and keep the exact value even when halving does not give a whole number. Working: the term before the one wanted is 25, and halving it means working out 25 ÷ 2, which is 12 with 1 left over to share, giving a half. Answer: 12.5. The distractors: 12 comes from halving 25 and then cutting the result down to a whole number; 0 comes from treating the sequence as one with a constant difference and taking 25 away from 25; 6.25 comes from halving twice and giving the term after the next one.
- (c) 70 — Method: the numbers of tiles form a sequence in which the same amount is added for each extra row, so the total for a number of rows is that amount multiplied by the number of rows. Working: 20 − 10 = 10 and 30 − 20 = 10, so each row adds 10 tiles; 7 rows therefore need 7 lots of 10, that is 7 × 10. Answer: 70. The distractors: 80 comes from counting one row too many and giving the total for 8 rows; 17 comes from adding the 10 tiles to the 7 rows instead of multiplying; 10 comes from giving the number of tiles in a single row rather than the total for all the rows.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (b) 14 — Substitute n=4 into 3n+2: 3×4+2=14. A candidate who adds 3 and n instead of multiplying would compute 3+4+2=9. A candidate who substitutes the wrong term number, n=3, would reach 3×3+2=11. A candidate who forgets to add the constant term would compute just 3×4=12.
- (b) 28 — Method: find the common difference, then use the position-to-term rule (or extend the sequence) to reach the 9th term. Working: the common difference is 7 − 4 = 3, so the nth term is 4 + 3(n − 1). For n = 9: 4 + 3 × 8 = 4 + 24 = 28. Answer: 28. 27 comes from using 3n instead of 3n + 1, dropping the constant term from the rule, 3 × 9 = 27. 31 comes from extending from the 5th term by one step too many, adding 3 five times instead of four, 16 + 3 × 5. 25 comes from extending by one step too few, adding 3 three times instead of four, 16 + 3 × 3.
- (a) 11 — Method: work out each term separately using the rule n² − 3, then subtract. Working: 6th term = 6² − 3 = 36 − 3 = 33. 5th term = 5² − 3 = 25 − 3 = 22. Difference: 33 − 22 = 11. Answer: 11. 8 comes from subtracting the constant −3 once at the end instead of it already being included in both terms, (36 − 25) − 3. 1 comes from working out (6 − 5)² instead of finding 6² and 5² separately and then subtracting. −11 comes from subtracting in the wrong order, the 5th term minus the 6th term instead of the 6th minus the 5th.
- (b) 21 — Method: find the gap between neighbouring terms, then add one gap to the last term that is known. Working: 9 − 5 = 4, 13 − 9 = 4 and 17 − 13 = 4, so 4 is added each time; the 5th term is one step on from the 4th term, so it is 17 + 4. Answer: 21. The distractors: 25 comes from adding the 4 twice and landing on the 6th term; 20 comes from multiplying the position by the common difference, 5 × 4, and ignoring the fact that the sequence starts at 5 rather than at 4; 22 comes from adding the first term, 5, to 17 instead of adding the common difference.
- (c) 100 — Method: apply the position-to-term rule straight to the position asked for, since it does not depend on the term before it. Working: the rule squares the position number and the position is 10, so the term is 10², which means 10 × 10. Answer: 100. The distractors: 81 comes from squaring 9 instead of 10, one position short; 20 comes from multiplying the position by 2 instead of squaring it; 1000 comes from cubing the position, 10 × 10 × 10, instead of squaring it.
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