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 − 3, where n is the position number. Work out the difference between the 6th term and the 5th term.
- 2.The nth term of a sequence is 3n + 2. Work out the 4th term of the sequence.
- 3.A sequence has the position-to-term rule n² + 2, where n is the position number. Work out the 6th term.
- 4.Here are the first five terms of an arithmetic sequence: 4, 7, 10, 13, 16. Work out the 9th term.
- 5.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.
- 6.Here are the first four terms of an arithmetic sequence: 5, 9, 13, 17. Work out the 5th term.
- 7.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.
- 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.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.
- 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.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.
- 12.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.
- 13.Which of these rules generates the sequence 6, 11, 16, 21, …?
- 14.Here are the first five terms of an arithmetic sequence: 2, 5, 8, 11, 14. Work out the 8th term.
- 15.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?
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) 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.
- (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) 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) 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) 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) 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.
- (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) 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.
- (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.
- (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) 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) 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.
- (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) 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.
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