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.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.
- 2.Here are the first four terms of an arithmetic sequence: 5, 9, 13, 17. Work out the 5th term.
- 3.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.
- 4.A theatre's front row has 18 seats. Each row behind has 4 more seats than the row in front. Which row has exactly 62 seats?
- 5.A geometric sequence has first term 5, and each term after that is 3 times the term before it. Work out the first term of the sequence that is greater than 1000.
- 6.Write down the first four terms of the sequence with nth term 6n − 5.
- 7.A geometric sequence starts at 3, and each term after that is found by multiplying the term before it by 2. Write down the first four terms.
- 8.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.
- 9.Matchsticks are laid out as a row of squares, with each new square sharing a side with the square before it. The first square uses 4 matchsticks and every extra square uses 3 more. Work out how many matchsticks a row of 4 squares uses.
- 10.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.
- 11.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 12.Here are the first five terms of an arithmetic sequence: 2, 5, 8, 11, 14. Work out the 8th term.
- 13.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 14.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.
- 15.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.
Answer key
- (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.
- (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.
- (d) 12 — Method: write the nth term of the sequence, 18 + 4(n − 1), set it equal to 62, and solve for n. Working: 18 + 4(n − 1) = 62, so 4(n − 1) = 44, giving n − 1 = 11, so n = 12. Answer: row 12. 11 comes from using 18 + 4n = 62 instead of 18 + 4(n − 1) = 62, an off-by-one error, giving n = 11. 48 comes from correctly simplifying to 4n = 48 but stopping there, without dividing by 4 to find n. 15.5 comes from dividing 62 by 4 directly, ignoring the 18 seats already in the front row.
- (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.
- (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.
- (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) 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) 13 — Method: the numbers of matchsticks form a sequence with a term-to-term rule, so count the first square in full and then add the repeated amount once for every extra square. Working: one square uses 4 matchsticks; a row of 4 squares has 3 extra squares after the first, and each of those adds 3 matchsticks, giving 3 × 3 = 9 to add on to the 4. Answer: 13. The distractors: 16 comes from counting each square as a separate set of 4 matchsticks, 4 × 4, and ignoring the shared sides; 12 comes from using 3 matchsticks for all four squares, 3 × 4, and forgetting that the first square needs a fourth side; 10 comes from adding the 3 only twice, as though a row of four squares had two extra squares rather than three.
- (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) 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) 23 — Method: find the number added each time, then add it to the first term once for every step between the first term and the term wanted. Working: 5 − 2 = 3, 8 − 5 = 3 and so on, so 3 is added each time; the 8th term is seven steps on from the 1st term, so it is 2 + 7 × 3 = 2 + 21. Answer: 23, which agrees with counting on from the 5th term: 14, 17, 20 and then one step more. The distractors: 26 comes from adding the common difference eight times instead of seven, 2 + 8 × 3; 20 comes from stopping one term early, at the 7th term; 24 comes from multiplying the position by the common difference, 8 × 3, and ignoring the fact that the sequence starts at 2 rather than at 3.
- (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.
- (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) 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.
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