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.
Answer key: Generating sequences worksheet — GCSE Foundation
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- (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.
- (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.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (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.
- (c) 5n + 1 — Method: find the common difference, then use it as the coefficient of n in the position-to-term rule, and find the constant by checking against the first term. Working: the common difference is 5, so the rule has the form 5n + c. Using the 1st term: 5(1) + c = 6, so c = 1. The rule is 5n + 1. Answer: 5n + 1. 5n − 1 uses the correct coefficient but the wrong sign for the constant. 6n comes from using the first term as the coefficient of n instead of the common difference — it matches the 1st term by coincidence but fails from the 2nd term onward. n + 5 swaps the coefficient and the constant around, using the common difference as the constant instead of the coefficient of n.
- (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.
- (b) 4, 12, 36, 108 — Method: multiply the previous term by 3 each time, starting from the first term. Working: 4 × 3 = 12, 12 × 3 = 36, 36 × 3 = 108. Answer: 4, 12, 36, 108. 4, 7, 10, 13 comes from adding 3 each time instead of multiplying by 3, confusing this with an arithmetic sequence. 4, 12, 15, 18 comes from multiplying correctly to get the second term, then switching to adding 3 for the rest. 12, 36, 108, 324 comes from listing the terms after the first term, missing off the starting value of 4.
- (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.
- (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.
- (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.
- (d) 3, 6, 12, 24 — Method: a term-to-term rule is applied to the term just written, so start with the first term and use the rule three more times. Working: the first term is 3; 3 × 2 = 6 is the second; 6 × 2 = 12 is the third; 12 × 2 = 24 is the fourth. Answer: 3, 6, 12, 24. The distractors: 3, 5, 7, 9 comes from adding 2 each time instead of multiplying by 2; 3, 9, 27, 81 comes from multiplying by the first term, 3, instead of by the multiplier 2 the rule gives; 6, 12, 24, 48 comes from doubling before writing anything down, so the list starts one term too late and the given first term is missing.
- (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) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (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.
- (b) 25 — Substitute n = 7: 4 × 7 − 3 = 28 − 3 = 25. Forgetting to subtract 3 gives 4 × 7 = 28. Subtracting 3 from 7 before multiplying by 4, 4 × (7 − 3) = 16, applies the operations in the wrong order. Substituting n = 8 by miscounting the position gives 4 × 8 − 3 = 29.
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