Cambridge International AS & A Level Mathematics 9709: Series – 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 1.6, Series, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027: the binomial expansion of (a + b)ⁿ for a positive integer n, arithmetic and geometric progressions, and the sum to infinity of a convergent geometric progression. It is AS Level content from Pure Mathematics 1, examined on Paper 1. A scientific calculator is allowed in all 9709 papers, but answers without working earn no marks, so show every step. Give non-exact answers to 3 significant figures.

Revise first with the study guide and the revision notes. The 9709 course hub and the printable checklist show where this section sits, the free 9709 AS diagnostic checks your AS topics together, and the 9709 self-check bank has more short questions.

Questions

1. Find the first three terms, in ascending powers of x, of (2 + x/4)⁸. Simplify each coefficient. [3]

2. For each sequence, state whether it is an arithmetic progression, a geometric progression or neither. Give the common difference or common ratio where there is one.

(a) 7, 4, 1, −2, … [1] (b) 12, −6, 3, −1.5, … [1] (c) 2, 6, 12, 20, … [1]

3. Find the term independent of x in the expansion of (x/2 + 4/x³)⁸. [3]

4. (a) Find the first three terms, in ascending powers of x, of (1 + kx)⁷, where k is a constant. [2] (b) The coefficient of x² in the expansion of (2 − x)(1 + kx)⁷ is 35. Find the possible values of k. [3]

5. Amira is training for a race. She runs 3 km on day 1, and on each later day she runs 0.4 km more than on the day before.

(a) Find the distance she runs on day 15. [2] (b) Find the total distance she runs in the first 30 days. [2] (c) Find the day on which her total distance first exceeds 200 km. [3]

6. A geometric progression has first term 6 and common ratio (x − 1)/2.

(a) Find the set of values of x for which the progression is convergent. [2] (b) The sum to infinity of the progression is 8. Find x. [2]

7. The numbers k − 3, k + 1 and 4k − 2 are the first three terms of a geometric progression.

(a) Show that 3k² − 16k + 5 = 0. [2] (b) Find the two possible values of k. [2] (c) For one value of k the progression is convergent. Find its sum to infinity. [3]

8. An arithmetic progression has first term 4 and common difference d, where d ≠ 0. The first, third and eleventh terms of the arithmetic progression are the first three terms of a geometric progression.

(a) Show that d = 6. [3] (b) State the common ratio of the geometric progression. [1] (c) Find the sum of the first 20 terms of the arithmetic progression. [2] (d) Find the sum of the first 6 terms of the geometric progression. [2]

9. A workshop has two machines. Machine A makes 500 parts in week 1, and each week after that it makes 96% of the number it made the week before. Machine B makes 300 parts in week 1, and each week after that it makes 12 more than the week before.

(a) Find the number of parts machine A makes in week 10. [2] (b) Find the total number of parts machine A makes in the first 20 weeks. [2] (c) Show that machine A can never make more than 12 500 parts in total. [2] (d) Find how many more parts machine B makes than machine A in the first 20 weeks. [3]

10. (a) Find the constant term and the term in 1/x² in the expansion of (x − 2/x)⁶. [3] (b) Hence find the term independent of x in the expansion of (x² + 3)(x − 2/x)⁶. [2]

11. The sum of the first n terms of a progression is given by Sₙ = 2n² + 5n.

(a) Find the first term and the second term. [2] (b) Show that the nth term is 4n + 3, and hence explain why the progression is arithmetic. [3] (c) Find the sum of the 21st to the 40th terms inclusive. [2]

12. A convergent geometric progression has first term a and common ratio r. Its sum to infinity is 4/3 of the sum of its first two terms.

(a) Show that r² = ¼. [3] (b) Given also that r > 0 and the sum of the first three terms is 21, find a and the sum to infinity. [3]

Answers

1. Terms of the form ⁸Cᵣ × 2⁸⁻ʳ × (x/4)ʳ [1]. 2⁸ = 256 and ⁸C₁ × 2⁷ × x/4 = 256x [1]. ⁸C₂ × 2⁶ × x²/16 = 28 × 4x² = 112x², so 256 + 256x + 112x² [1]. Examiner insight: the x² term is credited separately from the first two terms, so an error in it does not cost you the other two marks.

2. (a) Differences are all −3: AP, d = −3 [1]. (b) Ratios are all −½: GP, r = −½ [1]. (c) Differences 4, 6, 8 change and ratios 3, 2, 5/3 change: neither [1]. Examiner insight: a bare “GP” with no ratio does not earn the mark when the question asks for r.

3. General term ⁸Cᵣ (x/2)⁸⁻ʳ (4/x³)ʳ has power of x equal to 8 − r − 3r = 8 − 4r [1]. For no x, 8 − 4r = 0, so r = 2 [1]. Term = ⁸C₂ × (½)⁶ × 4² = 28 × 16/64 = 7 [1]. Examiner insight: the method mark needs a correct power of x for the general term; guessing r without it risks losing all three marks.

4. (a) 1 + 7kx [1] + 21k²x² [1], so 1 + 7kx + 21k²x². (b) Coefficient of x² is 2 × 21k² + (−1) × 7k = 42k² − 7k [1]. 42k² − 7k = 35 gives 6k² − k − 5 = 0, so (6k + 5)(k − 1) = 0 [1]. k = 1 or k = −5/6 [1]. Examiner insight: the method mark in (b) needs both products; using only 2 × 21k² is the most common way to lose it.

5. (a) a = 3, d = 0.4, u₁₅ = 3 + 14 × 0.4 [1] = 8.6 km [1]. (b) S₃₀ = ½ × 30 × (6 + 29 × 0.4) [1] = 15 × 17.6 = 264 km [1]. (c) Sₙ = ½n(6 + 0.4(n − 1)) = n(n + 14)/5, so n(n + 14)/5 > 200 gives n² + 14n − 1000 > 0 [1]. The positive root is n = (−14 + √4196)/2 ≈ 25.4 [1]. S₂₅ = 195 and S₂₆ = 208, so day 26 [1]. Examiner insight: the final mark needs a whole-number day; 25.4 or “day 25” loses it even with correct working.

6. (a) |(x − 1)/2| < 1, so −2 < x − 1 < 2 [1], giving −1 < x < 3 [1]. (b) 6/(1 − r) = 8 gives r = ¼ [1], so (x − 1)/2 = ¼ and x = 3/2, which lies in the range from (a) [1]. Examiner insight: in (a) the answer must be a strict inequality; writing −1 ≤ x ≤ 3 loses the accuracy mark.

7. (a) (k + 1)² = (k − 3)(4k − 2) [1]. So k² + 2k + 1 = 4k² − 14k + 6, which rearranges to 3k² − 16k + 5 = 0 [1]. (b) (3k − 1)(k − 5) = 0 [1], so k = 1/3 or k = 5 [1]. (c) k = 5 gives 2, 6, 18 with r = 3, which is not convergent. k = 1/3 gives −8/3, 4/3, −2/3 [1] with r = −½ [1]. S∞ = (−8/3)/(1 + ½) = −16/9 [1]. Examiner insight: (a) is a “show that” with the answer given, so the expanded line must appear before the printed equation.

8. (a) The terms are 4, 4 + 2d and 4 + 10d. Using b² = ac: (4 + 2d)² = 4(4 + 10d) [1]. 16 + 16d + 4d² = 16 + 40d, so 4d² − 24d = 0 [1]. 4d(d − 6) = 0 and d ≠ 0, so d = 6 [1]. (b) Terms 4, 16, 64, so r = 4 [1]. (c) S₂₀ = ½ × 20 × (8 + 19 × 6) [1] = 1220 [1]. (d) S₆ = 4(4⁶ − 1)/(4 − 1) [1] = 5460 [1]. Examiner insight: to earn the last mark in (a) you must reject d = 0 with a reason, because the answer is given.

9. (a) GP with a = 500, r = 0.96: u₁₀ = 500 × 0.96⁹ [1] = 346.3, so 346 parts [1]. (b) S₂₀ = 500(1 − 0.96²⁰)/(1 − 0.96) [1] = 6970 parts (3 s.f.) [1]. (c) |r| = 0.96 < 1, so the progression converges [1]. S∞ = 500/(1 − 0.96) = 12 500, and every partial sum is less than this, so the total never exceeds 12 500 [1]. (d) B is an AP with a = 300, d = 12: S₂₀ = ½ × 20 × (600 + 19 × 12) = 8280 [1]. Difference = 8280 − 6974.97 [1] = 1310 parts (3 s.f.) [1]. Examiner insight: in (c) a numerical total such as S₁₀₀ is not a proof; the mark needs |r| < 1 and S∞ = 12 500 stated.

10. (a) General term ⁶Cᵣ x⁶⁻ʳ (−2/x)ʳ = ⁶Cᵣ (−2)ʳ x⁶⁻²ʳ [1]. Constant: r = 3, ⁶C₃ × (−8) = −160 [1]. Term in 1/x²: r = 4, ⁶C₄ × 16 = 240/x² [1]. (b) x² × 240/x² + 3 × (−160) [1] = 240 − 480 = −240 [1]. Examiner insight: “Hence” means you must use the terms from (a); a fresh expansion of the product gains no credit if (a) is not used.

11. (a) u₁ = S₁ = 7 [1]. u₂ = S₂ − S₁ = 18 − 7 = 11 [1]. (b) uₙ = Sₙ − Sₙ₋₁ = 2n² + 5n − 2(n − 1)² − 5(n − 1) [1] = 4n + 3 [1]. Then uₙ₊₁ − uₙ = 4, a constant, so the progression is arithmetic with d = 4 [1]. (c) S₄₀ − S₂₀ = 3400 − 900 [1] = 2500 [1]. Examiner insight: in (b) the explanation mark needs the constant difference stated; showing the formula alone is not enough.

12. (a) a/(1 − r) = (4/3)a(1 + r) [1]. Divide by a (a ≠ 0): 3 = 4(1 − r)(1 + r) [1]. So 3 = 4(1 − r²), giving r² = ¼ [1]. (b) r = ½, so a(1 + ½ + ¼) = 21 [1], giving a = 12 [1]. S∞ = 12/(1 − ½) = 24 [1]. Examiner insight: in (a) the factorised or expanded 1 − r² step must be seen; going straight from the first equation to r² = ¼ loses marks on an answer-given part.

Where marks are usually lost

• Leaving the coefficient of b outside the power, for example writing x²/4 instead of x²/16 for (x/4)².
• Ignoring the negative sign in b, so odd powers of −2/x come out positive.
• Forgetting one of the two products when finding a coefficient in a product of brackets.
• Using d or r from the wrong term, for example u₁₁ = a + 11d.
• Giving a decimal or rounded-down number of terms or days.
• Quoting a sum to infinity for a ratio with |r| ≥ 1, or keeping both values of k when only one gives a convergent GP.
• Rounding mid-calculation, then losing the accuracy mark on a 3 s.f. answer.
• Skipping algebra on “show that” parts, where the answer is printed and every step must be visible.
• Dividing by a variable that could be zero (d or a) without saying why it is non-zero.

Next steps

• Series revision notes
• Series study guide
• Pure Mathematics 1 mixed practice for series alongside other Paper 1 topics
• 9709 course hub and printable checklist
• 9709 AS diagnostic and all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027, Version 4, Cambridge Assessment International Education (part of Cambridge University Press & Assessment). Section 1.6, Series (Pure Mathematics 1, for Paper 1).