Printable · GCSE Higher · ages 14-16
Special sequences: square, cube, Fibonacci, quadratic, geometric worksheet — GCSE Higher
Fifteen questions on "special sequences: square, cube, fibonacci, quadratic, geometric" — DfE statement A24. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Special sequences: square, cube, Fibonacci, quadratic, geometric worksheet — GCSE Higher
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- 1.Four sequences are shown below. Sequence P: 4, 8, 16, 32, ... Sequence Q: 4, 8, 12, 16, ... Sequence R: 4, 7, 12, 19, ... Sequence S: 4, 9, 16, 25, ... Work out which of the sequences P, Q, R and S is geometric.
- 2.The first term of an arithmetic sequence is 5, and each term after that increases by 6. Work out the 8th term of the sequence.
- 3.A geometric sequence has first term 3 and common ratio 2. Work out the 5th term of the sequence.
- 4.The first four terms of a sequence are 2, 2√3, 6, 6√3, ... Work out the next term.
- 5.A Fibonacci-type sequence begins 3, 5, 8, 13, ... where each term after the second is the sum of the two terms before it. Work out the 7th term of the sequence.
- 6.A sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₃ = 11 and a₅ = 29, work out a₁.
- 7.The first five terms of a quadratic sequence are 5, 8, 13, 20, 29. Work out the next term in the sequence.
- 8.A Fibonacci-type sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₁ = 5 and a₃ = 17, work out a₂.
- 9.A fundraising chain letter starts with 3 people taking part in round 1. Each following round has 4 times as many people taking part as the round before. Work out the number of people taking part in round 4.
- 10.A ball is dropped and bounces. The height of each bounce after the first is 8 cm less than the bounce before it. The first bounce reaches 60 cm. Work out the height of the 5th bounce.
- 11.Four sequences are shown below. Sequence P: 3, 6, 12, 24, … Sequence Q: 1, 4, 9, 16, … Sequence R: 2, 4, 8, 14, … Sequence S: 5, 10, 15, 20, … Work out which of the sequences P, Q, R and S is geometric.
- 12.A geometric sequence begins 80, 40, 20, 10, ... Work out the next term in the sequence.
- 13.A geometric sequence begins 162, 54, 18, 6, ... Work out the next term in the sequence.
- 14.A company's profit is £500 in its first year. Each year after that, the profit is £300 more than the year before. Work out the profit in the company's 6th year.
- 15.An arithmetic sequence has first term 40 and common difference −6. Work out the 9th term of the sequence.
Answer key
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (a) 18 — The common ratio is 2√3 ÷ 2 = √3. Checking: 6 ÷ 2√3 = √3 and 6√3 ÷ 6 = √3, so the ratio is consistent throughout. The next term is 6√3 × √3 = 6 × 3 = 18, since √3 × √3 = 3. Looking only at the coefficients 2, 2, 6, 6 and continuing them by doubling the last one gives 6 × 2 = 12, which is wrong because the step from each term to the next is a multiplication by √3, not a pattern in the coefficients alone. Doubling the previous term instead of multiplying by the surd ratio √3 gives 6√3 × 2 = 12√3, which is wrong because the common ratio is √3, not 2. Using 3 instead of √3 as the common ratio — squaring the true ratio by mistake — gives 6√3 × 3 = 18√3, which is wrong because 3 is the SQUARE of the common ratio, not the ratio itself.
- (c) 55 — Continuing the pattern: 8+13=21 (5th term), 13+21=34 (6th term), 21+34=55 (7th term). A candidate who miscounts the position and stops one term early would give 34, the 6th term. A candidate who doubles the most recent term instead of adding the two before it would compute 34×2=68. A candidate who adds a non-adjacent pair — the 4th and 6th terms, skipping the 5th — would compute 13+34=47.
- (a) 4 — a₅ = a₄ + a₃, so a₄ = a₅ − a₃ = 29 − 11 = 18. a₄ = a₃ + a₂, so a₂ = a₄ − a₃ = 18 − 11 = 7. a₃ = a₂ + a₁, so a₁ = a₃ − a₂ = 11 − 7 = 4. Checking forwards: 4, 7, 11, 18, 29. Answering 7 stops one step early, reporting a₂ = 7 instead of continuing back one more step to a₁ — wrong, because the question asks for a₁, not a₂. Answering 18 reports a₄ = 18, an intermediate value found along the way, instead of a₁ — wrong, because a₄ is a term used to reach the answer, not the term the question asks for. Answering 3 takes one backward step too many, working out a further term a₀ = a₂ − a₁ = 7 − 4 = 3 — wrong, because the sequence starts at a₁, so a₁ = 4 is as far back as the question goes.
- (b) 40 — The differences between consecutive terms are 3, 5, 7, 9, increasing by 2 each time, so the next difference is 11, giving 29 + 11 = 40. A candidate who reuses the last difference of 9 without increasing it gets 29 + 9 = 38. A candidate who increases the difference by only 1 instead of 2 gets 29 + 10 = 39. A candidate who reuses the first difference of 3 instead of the pattern gets 29 + 3 = 32.
- (d) 12 — Since a₃ = a₂ + a₁, rearranging gives a₂ = a₃ − a₁ = 17 − 5 = 12. 22 comes from adding instead of subtracting: a₃ + a₁ = 17 + 5 = 22. 11 comes from treating the sequence as if it were arithmetic and averaging the two given terms: (5 + 17) ÷ 2 = 11 — but a Fibonacci-type sequence isn't arithmetic, so this doesn't apply. −12 comes from subtracting in the wrong order: a₁ − a₃ = 5 − 17 = −12.
- (b) 192 — This is a geometric sequence with first term 3 and common ratio 4: round 2 has 3 × 4 = 12, round 3 has 12 × 4 = 48, round 4 has 48 × 4 = 192. A candidate who applies the ×4 multiplier four times instead of three gets 3 × 4⁴ = 768. A candidate who wrongly treats the growth as arithmetic, taking the round 2 figure of 12 as a fixed amount added each round, gets 3, 15, 27, 39. A candidate who forgets the starting 3 people and just works out 4⁴ gets 256.
- (d) 28 — The height decreases by 8 cm at each bounce after the first, so the nth bounce reaches 60−(n−1)×8 cm. For the 5th bounce: 60−4×8=60−32=28. A candidate who subtracts 8 one time too many, five times instead of four, would compute 60−5×8=20. A candidate who adds the decrease instead of subtracting it, a sign error, would compute 60+4×8=92. A candidate who works out only the total decrease and forgets to include the starting height of 60 cm would compute just 5×8=40.
- (a) P — A sequence is geometric when consecutive terms share a constant ratio. For P: 6 ÷ 3 = 2, 12 ÷ 6 = 2, 24 ÷ 12 = 2 — the ratio is constant at 2, so P is geometric. Q is the square numbers (1², 2², 3², 4²), a quadratic sequence: its ratios are 4, 2.25, 1.78, … — not constant. R looks geometric at first (2, 4, 8 doubles each time), but the pattern breaks: 8 to 14 is a ratio of 1.75, not 2. Its first differences are 2, 4, 6 — a constant second difference of 2 — so R is a quadratic sequence, not geometric. S has a constant DIFFERENCE of 5 (it is arithmetic), but its ratios (2, 1.5, 1.33, …) are not constant, so it is not geometric.
- (d) 5 — Each term is found by multiplying the previous term by the common ratio, 0.5: 80, 40, 20, 10, and the next term is 10 × 0.5 = 5. A candidate who instead subtracts the same amount each time (repeating the last difference of 10) would reach 10 − 10 = 0. A candidate who divides by 4 instead of by 2 would reach 10 ÷ 4 = 2.5. A candidate who multiplies by 2 instead of dividing (reversing the direction of the sequence) would reach 10 × 2 = 20.
- (a) 2 — The common ratio is 54 ÷ 162 = 1/3, confirmed by 18 ÷ 54 = 1/3 and 6 ÷ 18 = 1/3. The next term is 6 × 1/3 = 2. A candidate who wrongly treats the sequence as arithmetic and subtracts the last difference (18 − 6 = 12) again gets 6 − 12 = −6. A candidate who multiplies by 3 instead of dividing, effectively reversing the sequence, gets 18. A candidate who guesses the ratio is 1/2 gets 6 × 1/2 = 3.
- (d) £2000 — This is an arithmetic sequence with first term £500 and common difference £300. The 6th term is 500 + 5 × 300 = 2000. A candidate who uses 6 lots of the increase instead of 5 gets 500 + 6 × 300 = 2300. A candidate who forgets to add the first year's profit at all gets 5 × 300 = 1500. A candidate who miscounts the number of increases as 4 instead of 5 gets 500 + 4 × 300 = 1700.
- (c) −8 — The nth term is the first term plus (n − 1) lots of the common difference: 40 + 8 × (−6) = 40 − 48 = −8. A candidate who uses 9 lots of the common difference instead of 8 gets 40 + 9 × (−6) = −14. A candidate who treats the common difference as +6 instead of −6 gets 40 + 8 × 6 = 88. A candidate who uses only 7 lots of the common difference gets 40 + 7 × (−6) = −2.
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