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 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.
- 2.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.
- 3.A sequence has the position-to-term rule: the nth term is 3n. Write down the first four terms of the sequence.
- 4.The nth term of a sequence is 3n + 2. Work out the 4th 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 4n − 3. Work out the 7th term of the sequence.
- 7.The nth term of a sequence is 2n − 1. Work out the 6th term of the sequence.
- 8.A geometric sequence starts at 4, and each term after that is found by multiplying the term before it by 3. Write down the first four terms of the sequence.
- 9.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.
- 10.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.
- 11.Here are the first five terms of an arithmetic sequence: 2, 5, 8, 11, 14. Work out the 8th term.
- 12.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.
- 13.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.
- 14.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?
- 15.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
Answer key
- (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) 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) 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) 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) 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.
- (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.
- (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) 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.
- (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.
- (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.
- (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.
- (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) 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) 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.
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