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.
Generating sequences worksheet — GCSE Foundation
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- 1.A sequence begins at 60, and each term after that is found by subtracting 7 from the term before it. Work out the 5th term of the sequence.
- 2.The nth term of a sequence is 2n² + 1. Work out the 4th term of the sequence.
- 3.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.
- 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 nth term 3n + 4. Two terms are missing from the list of its first five terms: 7, 10, _, 16, _. Work out the missing 3rd and 5th terms added together.
- 6.Write down the first four terms of the sequence with nth term 6n − 5.
- 7.A plant is 4 cm tall in week 1. Each week after that, its height increases by 5 cm. Work out the first week in which the plant is taller than 30 cm.
- 8.A sequence has the position-to-term rule 5n − 2, where n is the position number. Which of these is a term in the sequence?
- 9.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 10.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 11.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?
- 12.A sequence has the position-to-term rule: the nth term is 3n. Write down the first four terms of the sequence.
- 13.A sequence has the position-to-term rule n² + 2, where n is the position number. Work out the 6th term.
- 14.Here are the first four terms of an arithmetic sequence: 5, 9, 13, 17. Work out the 5th term.
- 15.The nth term of a sequence is 3n + 2. Work out the 4th term of the sequence.
Answer key
- (c) 32 — The terms are 60, 53, 46, 39, 32 — each found by subtracting 7 from the term before, so the 5th term is 32. Subtracting 7 five times from the first term instead of four times, 60 − 7 × 5 = 25, treats the first term as if it came before the sequence starts. Adding 7 four times instead of subtracting, 60 + 7 × 4 = 88, uses the wrong operation. Stopping one term early gives the 4th term, 39.
- (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.
- (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.
- (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.
- (c) 32 — Using the rule 3n + 4: the 3rd term is 3 × 3 + 4 = 13, and the 5th term is 3 × 5 + 4 = 19, so their sum is 13 + 19 = 32. Forgetting to add the 4 for the 3rd term, 3 × 3 = 9, and adding the correct 5th term, gives 9 + 19 = 28. Rounding the 19 up to 20 to make the addition easier and then forgetting to take the extra 1 back off, 13 + 20 = 33, gives 33. Using the rule 4n + 3 instead of 3n + 4 gives 4 × 3 + 3 = 15 and 4 × 5 + 3 = 23, summing to 38.
- (c) 1, 7, 13, 19 — Method: substitute n = 1, 2, 3, 4 into the rule 6n − 5 in turn. Working: n = 1: 6 − 5 = 1. n = 2: 12 − 5 = 7. n = 3: 18 − 5 = 13. n = 4: 24 − 5 = 19. Answer: 1, 7, 13, 19. 6, 12, 18, 24 comes from using 6n on its own, forgetting to subtract 5. 5, 11, 17, 23 comes from using the rule 6n − 1 instead of 6n − 5, a slip in the constant. 0, 6, 12, 18 comes from using 6(n − 1) instead of 6n − 5, effectively shifting every term one position along.
- (b) Week 7 — The height after n weeks is 4 + 5(n − 1) = 5n − 1. The plant is taller than 30 cm when its height, 5n − 1, is more than 30, written 5n − 1 > 30; add 1 to both sides (5n > 31) and divide by 5, which gives n > 6.2, so the first whole week is week 7, where the height is 5 × 7 − 1 = 34 cm. Rounding 6.2 down to week 6 without checking gives a height of only 5 × 6 − 1 = 29 cm, which is not yet taller than 30 cm. Counting one week too many gives week 8. Working out 30 ÷ 5 = 6 and then subtracting 1 from the week number instead of from the height gives week 5.
- (c) 28 — Method: substitute values of n into 5n − 2 and check which of the options matches. Working: for n = 6, 5 × 6 − 2 = 30 − 2 = 28, so 28 is a term of the sequence, the 6th term. Answer: 28. 27 comes from using the rule 5n − 3 instead of 5n − 2. 30 comes from using 5n on its own, forgetting to subtract 2 at all. 20 comes from applying the subtraction before the multiplication, working out 5 × (n − 2) instead of 5n − 2.
- (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 — 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.
- (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) 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.
- (a) 38 — Method: substitute the position number into the rule and follow the order of operations, so the squaring is carried out before the 2 is added. Working: n = 6 gives 6² + 2; 6² means 6 × 6 = 36, and then 2 is added to 36. Answer: 38. The distractors: 14 comes from multiplying the position by 2 instead of squaring it, 6 × 2 + 2; 36 comes from squaring correctly and then forgetting to add the 2; 64 comes from adding the 2 first and squaring afterwards, (6 + 2)².
- (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.
- (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.
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