AQA A-Level Mathematics: D: Sequences and series (7357) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover Section D: Sequences and series (D1 to D6) of AQA A-Level Mathematics (7357), from the AQA A-level Mathematics specification, version 1.3 (31 January 2018), for exams from June 2018 onwards. The specification lists Section D under Paper 1, and Papers 2 and 3 can assess any Paper 1 content. A calculator is required in every 7357 paper, so no question here is calculator-free, but each answer shows the algebra you are expected to write down.

Learn the content first with the Section D study guide and the Section D revision notes. Course links: AQA A-Level Mathematics hub and printable 7357 checklist.

Questions

1. Find the coefficient of x⁴ in the expansion of (2 − 3x)⁷. [3]

2. Calculate the value of Σ from r = 1 to 30 of (4r − 7). [3]

3. The random variable X has the distribution B(10, 0.2).

(a) State the term in the expansion of (0.8 + 0.2)¹⁰ that equals P(X = 3). [1] (b) Calculate P(X = 3), giving your answer to three significant figures. [2]

4. A sequence is defined by u₁ = 3 and uₙ₊₁ = (uₙ + 1)/(uₙ − 1).

(a) Find u₂, u₃ and u₄. [2] (b) Hence find the value of Σ from r = 1 to 50 of uᵣ. [2]

5. An arithmetic series is 5 + 9 + 13 + …

(a) Show that the sum of the first n terms is n(2n + 3). [2] (b) Find the least value of n for which the sum of the first n terms exceeds 2000. [3]

6.

(a) Find the binomial expansion of (1 − x)⁻² up to and including the term in x³. [2] (b) Hence find the expansion of (1 + 2x)/(1 − x)² up to and including the term in x³, and state the range of values of x for which it is valid. [3]

7. The infinite series Σ from r = 0 to ∞ of ((2x + 1)/3)ʳ is geometric.

(a) Find the set of values of x for which the series converges. [3] (b) Find the value of x for which the sum to infinity is 4. [2]

8. The second term of a geometric series is 24 and the fifth term is −3.

(a) Find the common ratio and the first term. [3] (b) Explain why the series has a sum to infinity, and find it. [2] (c) Find the exact value of S₁₀. [2]

9. In the expansion of (1 + kx)ⁿ, where n is a positive integer and k > 0, the coefficient of x is 20 and the coefficient of x² is 150.

(a) Find n and k. [4] (b) Find the coefficient of x³. [2]

10.

(a) Find the binomial expansion of (8 + 3x)^(1/3) in ascending powers of x up to and including the term in x². [4] (b) State the range of values of x for which the expansion is valid. [1] (c) Use your expansion with a suitable value of x to estimate the cube root of 8.3, giving your answer to five decimal places. [2]

11. Two monthly savings plans both start with a £200 deposit in month 1. In Plan A each deposit is £15 more than the one before. In Plan B each deposit is 4% more than the one before.

(a) Find the deposit in month 12 for each plan. [2] (b) Find the total deposited over the first 24 months for each plan, to the nearest penny. [3] (c) Plan B has a maximum deposit of £500. Find the first month in which the Plan B formula would give a deposit above £500. [3] (d) Give one reason why Plan B may not be a realistic model over many years. [1]

12. A patient takes a medicine once a day. A model for the amount, xₙ mg, in the body just after the nth dose is x₁ = 10 and xₙ₊₁ = 0.8xₙ + 6.

(a) Find x₂ and x₃. [2] (b) Let yₙ = xₙ − 30. Show that yₙ₊₁ = 0.8yₙ, and hence show that xₙ = 30 − 20(0.8)ⁿ⁻¹. [3] (c) State whether the sequence is increasing or decreasing, and state the long-term amount predicted by the model. [2] (d) Find the least n for which xₙ > 29. [2]

Answers

1. The x⁴ term is ⁷C₄ (2)³ (−3x)⁴. [1] Coefficient = 35 × 8 × 81 [1] = 22680. [1] [3] Examiner insight: Bracketing (−3x)⁴ is the method mark here; if you write −3x⁴ the sign and size both go wrong and the accuracy mark is lost.

2. The terms run from −3 (r = 1) to 113 (r = 30): an arithmetic series of 30 terms. [1] S = (30/2)(−3 + 113) [1] = 1650. [1] [3] Examiner insight: A method mark needs the correct number of terms; if you split into 4Σr − Σ7, write Σ7 as 7 × 30 = 210, not 7.

3. (a) ¹⁰C₃ (0.8)⁷ (0.2)³ [1] (b) 120 × 0.2097152 × 0.008 [1] = 0.201 [1] [3] Examiner insight: A bare calculator value gets the accuracy mark at best; showing ¹⁰C₃ pʳqⁿ⁻ʳ with the right numbers secures the method mark.

4. (a) u₂ = 4/2 = 2 [1]; u₃ = 3/1 = 3 and u₄ = 2. [1] (b) The sequence is periodic with order 2, and each pair sums to 5. [1] 50 terms make 25 pairs, so the sum is 125. [1] [4] Examiner insight: State the period explicitly before using it; a final answer with no reason given for grouping the terms risks losing the method mark.

5. (a) Sₙ = (n/2)(2 × 5 + (n − 1) × 4) [1] = (n/2)(4n + 6) = n(2n + 3), as required. [1] (b) 2n² + 3n − 2000 > 0. [1] The positive root is n = (−3 + √16009)/4 = 30.88… [1] S₃₀ = 1890 and S₃₁ = 2015, so n = 31. [1] [5] Examiner insight: On a “show that” every line must appear, including the simplification to (4n + 6); for (b), checking the two whole numbers either side shows why you rounded up.

6. (a) 1 + (−2)(−x) + (−2)(−3)(−x)²/2! + (−2)(−3)(−4)(−x)³/3! [1] = 1 + 2x + 3x² + 4x³ [1] (b) (1 + 2x)(1 + 2x + 3x² + 4x³) [1] = 1 + 4x + 7x² + 10x³ [1], valid for |x| < 1. [1] [5] Examiner insight: “Hence” means you must use part (a); multiplying out and dropping terms beyond x³ earns the method mark even if one coefficient slips.

7. (a) The common ratio is (2x + 1)/3. [1] Converges when −1 < (2x + 1)/3 < 1, so −3 < 2x + 1 < 3 [1], giving −2 < x < 1. [1] (b) 1/(1 − r) = 4 gives r = 3/4 [1], so 2x + 1 = 9/4 and x = 5/8, which is inside the range. [1] [5] Examiner insight: Give the range as a double inequality (or using modulus notation correctly); writing only x < 1 loses the final accuracy mark.

8. (a) ar = 24 and ar⁴ = −3. [1] Dividing, r³ = −1/8 [1], so r = −1/2 and a = −48. [1] (b) |r| = 1/2 < 1, so the series converges. [1] S∞ = −48/(1 + 1/2) = −32. [1] (c) S₁₀ = −48(1 − (−1/2)¹⁰)/(3/2) [1] = −32(1 − 1/1024) = −1023/32. [1] [7] Examiner insight: “Exact” means a fraction, not −31.97; and the reason in (b) must quote the condition |r| < 1, not just “r is small”.

9. (a) nk = 20 [1] and n(n − 1)k²/2 = 150. [1] Dividing, (n − 1)k = 15, so nk − k = 15 and k = 5 [1], then n = 4. [1] (b) ⁴C₃ × 5³ [1] = 500. [1] [6] Examiner insight: The method mark for the x² coefficient needs k² (from (kx)²); writing n(n − 1)k/2 instead leads to the wrong values n = 16 and k = 5/4.

10. (a) (8 + 3x)^(1/3) = 2(1 + 3x/8)^(1/3). [1] (1 + X)^(1/3) = 1 + X/3 − X²/9 + … [1] With X = 3x/8: 1 + x/8 − x²/64. [1] So the expansion is 2 + x/4 − x²/32. [1] (b) |3x/8| < 1, so |x| < 8/3. [1] (c) 8 + 3x = 8.3 gives x = 0.1: 2 + 0.025 − 0.0003125 [1] = 2.02469 (5 d.p.). [1] [7] Examiner insight: Forgetting to multiply by 8^(1/3) = 2 is a common slip that loses all the accuracy marks in (a); and the validity range must come from 3x/8, not from x.

11. (a) Plan A: 200 + 11 × 15 = £365. [1] Plan B: 200 × 1.04¹¹ = £307.89. [1] (b) Plan A: (24/2)(400 + 23 × 15) = £8940. [1] Plan B: 200(1.04²⁴ − 1)/0.04 [1] = £7816.52. [1] (c) 200 × 1.04ⁿ⁻¹ > 500, so 1.04ⁿ⁻¹ > 2.5. [1] n − 1 > ln 2.5/ln 1.04 = 23.36 [1], so n − 1 = 24 and the first such month is month 25 (month 24 gives £492.94, month 25 gives £512.66). [1] (d) For example, the saver’s income is unlikely to grow by 4% every month, so the deposits could not keep rising at that rate. [1] [9] Examiner insight: Being one term out (using 1.04ⁿ instead of 1.04ⁿ⁻¹) loses accuracy marks in (a) and (c); state which month you mean in words.

12. (a) x₂ = 0.8 × 10 + 6 = 14 [1]; x₃ = 0.8 × 14 + 6 = 17.2. [1] (b) yₙ₊₁ = xₙ₊₁ − 30 = 0.8xₙ − 24 = 0.8(xₙ − 30) = 0.8yₙ. [1] So y is geometric with y₁ = 10 − 30 = −20, giving yₙ = −20(0.8)ⁿ⁻¹. [1] Hence xₙ = 30 − 20(0.8)ⁿ⁻¹. [1] (c) 20(0.8)ⁿ⁻¹ decreases, so the sequence is increasing. [1] It tends to 30 mg. [1] (d) 20(0.8)ⁿ⁻¹ < 1, so n − 1 > ln 20/ln 1.25 = 13.43. [1] n = 15 (x₁₄ ≈ 28.90, x₁₅ ≈ 29.12). [1] [9] Examiner insight: In (b) you cannot just verify the formula for a few values of n; the marks need the general step showing yₙ₊₁ = 0.8yₙ.

Where marks are usually lost

• Writing (−3x)⁴ or (kx)² without brackets, so powers and signs are wrong.
• Forgetting to take out aⁿ when the constant in (a + bx)ⁿ is not 1, then giving the wrong validity range.
• Miscounting terms in sigma notation, or treating Σ of a constant as the constant itself.
• Rounding a “least n” answer down, or giving a decimal value of n.
• Not flipping the inequality after dividing by the log of a number between 0 and 1.
• Quoting S∞ without first stating |r| < 1.
• Using the wrong power in a geometric model (rⁿ for the nth term instead of rⁿ⁻¹).
• In periodic sequences, not stating the period before summing.
• Giving answers to models without units, months or context.

Next steps

• Recap with the Section D revision notes
• Relearn any weak part from the Section D study guide
• Practise modelling skills with the overarching themes practice
• Plan your papers with the 7357 exam preparation guide
• Browse the AQA A-Level Mathematics hub and tick off the 7357 checklist
• Try all free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for teaching from September 2017 and exams from June 2018 onwards, published by AQA. Section 3.5, D: Sequences and series (D1 to D6).