OxfordAQA International AS and A-level Mathematics: Sequences, series and binomial expansion (9660) – 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 P1.5 (Sequences and series) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International AS Unit P1. Each answer shows where marks would be earned.

Learn the content first from the study guide and revision notes.

Questions

1. A sequence has nth term x(n) = 3n² − 2. (a) Write down the first three terms. (b) Show that 145 is a term of the sequence and state which term it is. [3]

2. A sequence is defined by x(n+1) = 1/(1 − x(n)), with x(1) = 3. (a) Find x(2), x(3) and x(4). (b) Hence write down x(100), giving a reason. [4]

3. A sequence is defined by x(n+1) = (x(n)² + 6)/5, with x(1) = 0. (a) Find x(2) and x(3) as decimals. (b) Suppose the sequence converges to a limit L. Show that L satisfies L² − 5L + 6 = 0 and find the possible values of L. [5]

4. An arithmetic sequence has 4th term 19 and the sum of its first 12 terms is 348. (a) Find the first term and the common difference. [4] (b) Find the smallest term of the sequence that exceeds 200. [2]

5. (a) Evaluate Σ from r = 1 to 40 of (5 − 2r). [2] (b) Use the formula for the sum of the first n natural numbers to show that Σ from r = 1 to n of (2r − 1) = n². [3]

6. A geometric sequence has 3rd term 18 and 6th term 486. (a) Find the common ratio and the first term. [3] (b) Find the sum of the first 9 terms. [2] (c) Explain why the series does not have a sum to infinity. [1]

7. Maya pays £800 into a savings account at the start of each year for 10 years. The account grows by 4% each year, and nothing is withdrawn. (a) Show that the value at the end of the 10th year of the first payment is 800 × 1.04¹⁰. [1] (b) Show that the total value at the end of the 10th year is 800 × (1.04 + 1.04² + … + 1.04¹⁰), and hence find it to the nearest pound. [4]

8. A geometric series has first term 20 and sum to infinity 50. (a) Find the common ratio. [2] (b) Find the sum of all the terms from the 4th term onwards. [3]

9. By writing it as a geometric series, show that 0.2777… (with the 7 recurring) equals 5/18. [4]

10. A geometric series has first term 4 and common ratio (3x − 1)/2. (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 8. [2]

11. (a) Evaluate 8C3. [1] (b) Expand (3 + 2x)⁴, simplifying each coefficient. [4] (c) Hence find the coefficient of x² in the expansion of (2 + x)(3 + 2x)⁴. [2]

12. The coefficient of x³ in the expansion of (1 + kx)⁷ is 5 times the coefficient of x², where k is a non-zero constant. (a) Find k. [4] (b) Use the first three terms of the expansion of (1 + x)¹⁰, with x = 0.01, to estimate 1.01¹⁰. Given that 1.01¹⁰ = 1.104622 to 6 decimal places, find the error in your estimate. [3]

Answers

1. (a) x(1) = 3 − 2 = 1, x(2) = 12 − 2 = 10, x(3) = 27 − 2 = 25. [1] (b) 3n² − 2 = 145, so 3n² = 147 and n² = 49. [1] n = 7 (n = −7 is rejected because n is positive), so 145 is the 7th term. [1] Examiner insight: On a “show that” part, set the formula equal to 145 and solve, then comment that n is a positive whole number. Checking only that 3 × 49 − 2 = 145 leaves the position of the term unexplained.

2. (a) x(2) = 1/(1 − 3) = −1/2. [1] x(3) = 1/(1 + 1/2) = 2/3. [1] x(4) = 1/(1 − 2/3) = 3. [1] (b) x(4) = x(1), so the sequence repeats every 3 terms. 100 = 3 × 33 + 1, so x(100) = x(1) = 3. [1] Examiner insight: Keep fractions exact rather than using decimals, so that x(4) visibly returns to 3. A bare answer of 3 with no mention of the period is unlikely to earn the reason.

3. (a) x(2) = (0 + 6)/5 = 1.2. [1] x(3) = (1.44 + 6)/5 = 1.488. [1] (b) At the limit, x(n+1) and x(n) both equal L, so L = (L² + 6)/5. [1] Then 5L = L² + 6, so L² − 5L + 6 = 0. [1] (L − 2)(L − 3) = 0, so L = 2 or L = 3. [1] Examiner insight: Show the step where you replace both terms by L before rearranging. Give both roots unless the question tells you to choose one, because the equation alone does not decide which limit the sequence reaches.

4. (a) x(4) = a + 3d = 19. [1] S(12) = 6(2a + 11d) = 348, so 2a + 11d = 58. [1] Substituting a = 19 − 3d: 38 − 6d + 11d = 58, so d = 4. [1] Then a = 7, d = 4. [1] (b) 7 + 4(n − 1) > 200 gives n > 49.25, so n = 50. [1] The term is 7 + 4 × 49 = 203. [1] Examiner insight: Two unknowns need two equations, so write both clearly before eliminating. In part (b), test the whole number either side: the 49th term is 199, which does not exceed 200.

5. (a) The terms form an arithmetic series with first term 3, last term −75 and 40 terms. [1] S = 40/2 × (3 + (−75)) = 20 × (−72) = −1440. [1] (b) Σ(2r − 1) = 2Σr − Σ1 = 2 × n(n + 1)/2 − n. [1] = n(n + 1) − n = n² + n − n. [1] = n², as required. [1] Examiner insight: The last term of part (a) is 5 − 80 = −75, and many errors come from sign slips here. For “show that”, every line of the algebra should be visible, ending on the stated result.

6. (a) ar² = 18 and ar⁵ = 486. Dividing gives r³ = 27. [1] So r = 3 (the only real cube root). [1] a = 18/9 = 2. [1] (b) S(9) = 2(3⁹ − 1)/(3 − 1) = 3⁹ − 1. [1] = 19 683 − 1 = 19 682. [1] (c) The common ratio 3 does not satisfy |r| < 1, so the terms grow and the series does not converge. [1] Examiner insight: Dividing the two terms removes a and is much faster than guessing. State the condition |r| < 1 explicitly in part (c) and say that r = 3 breaks it.

7. (a) The payment is multiplied by 1.04 once for each of the 10 years, giving 800 × 1.04¹⁰. [1] (b) The first payment grows for 10 years, the second for 9, and so on, down to the last payment, which grows for 1 year. [1] The total is 800(1.04¹⁰ + 1.04⁹ + … + 1.04), which is the same sum written in reverse order. [1] This is geometric with first term 1.04 and ratio 1.04, so the total is 800 × 1.04(1.04¹⁰ − 1)/(1.04 − 1). [1] This equals 800 × 1.04 × 12.0061… = 9989.08, so £9989. [1] Examiner insight: Say how many years each payment grows, because that justifies the first term 1.04 rather than 1. Round only at the end: rounding 1.04¹⁰ early can shift the final pound.

8. (a) 20/(1 − r) = 50, so 1 − r = 0.4. [1] r = 0.6. [1] (b) The 4th term is 20 × 0.6³ = 4.32. [1] The sum from the 4th term onwards has first term 4.32 and ratio 0.6, so it equals 4.32/(1 − 0.6). [1] = 10.8. [1] Examiner insight: A second method is S∞ minus S(3) = 50 − 39.2 = 10.8. Either route is fine, but you need to state which terms you are summing. Check that 0.6 satisfies |r| < 1 before using the formula.

9. 0.2777… = 0.2 + 0.07 + 0.007 + 0.0007 + … [1] The terms after 0.2 form a geometric series with first term 0.07 and ratio 0.1. [1] Its sum to infinity is 0.07/(1 − 0.1) = 0.07/0.9 = 7/90. [1] Total: 2/10 + 7/90 = 18/90 + 7/90 = 25/90 = 5/18. [1] Examiner insight: Separate the non-repeating part (0.2) before forming the series. Starting the geometric series at 0.27 would give the wrong first term.

10. (a) Convergence needs |(3x − 1)/2| < 1, so |3x − 1| < 2. [1] −2 < 3x − 1 < 2. [1] −1 < 3x < 3, so −1/3 < x < 1. [1] (b) 4/(1 − r) = 8, so 1 − r = 1/2 and r = 1/2. [1] (3x − 1)/2 = 1/2 gives 3x − 1 = 1, so x = 2/3. [1] Examiner insight: Check that the x from part (b) lies in the interval from part (a). Here 2/3 does, which confirms the series really converges. A value outside the interval would be rejected.

11. (a) 8C3 = 8!/(3! 5!) = 8 × 7 × 6/6 = 56. [1] (b) Coefficients 1, 4, 6, 4, 1 with a = 3 and b = 2x. [1] Terms: 81, 4 × 27 × 2x = 216x, 6 × 9 × 4x² = 216x², 4 × 3 × 8x³ = 96x³, 16x⁴. [2] So (3 + 2x)⁴ = 81 + 216x + 216x² + 96x³ + 16x⁴. [1] (c) The x² terms come from 2 × 216x² and x × 216x. [1] 432 + 216 = 648. [1] Examiner insight: Numerical coefficients such as 2⁴ and 3² must be applied to every term. For part (c), list the two ways of reaching x² before adding, so a slip is easy to trace.

12. (a) The x² coefficient is 7C2 k² = 21k². [1] The x³ coefficient is 7C3 k³ = 35k³. [1] So 35k³ = 5 × 21k², giving 35k³ − 105k² = 0, i.e. 35k²(k − 3) = 0. [1] Since k is non-zero, k = 3. [1] (b) Estimate: 1 + 10(0.01) + 45(0.01)² = 1 + 0.1 + 0.0045 = 1.1045. [1] Error = 1.104622 − 1.1045 = 0.000122 (3 s.f.), so the estimate is slightly low. [2] Examiner insight: Do not divide through by k² without noting k is non-zero, as this is the condition that rules out the root k = 0. In part (b), state that the estimate is smaller than the true value.

Where marks are usually lost

• Using n instead of n − 1 as the power of r in the nth term of a geometric sequence.
• Not rejecting a root that cannot be a term number or limit, such as a negative n.
• Stating a sum to infinity without checking |r| < 1.
• Rounding n down in “least n” questions.
• Sign slips with negative terms, especially in (−3x)² and 5 − 2r.
• Forgetting to apply numerical coefficients to every term of a binomial expansion.
• Leaving out a way of reaching a power when expanding a product of brackets.
• Giving only one limit when L = f(L) has two roots and the question does not choose between them.

Next steps

• Review the formulae in the revision notes.
• Re-read the worked examples in the study guide.
• Check progress against the course hub and the checklist.
• Take the free 10-minute diagnostics to find the next topic to work on.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: section P1.5, Sequences and series, under International AS Unit P1 (Pure Maths).