Question Simulator

IGCSE Additional Mathematics (0606)

Series Banker

22 questions · about 13% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 12 · Q4 (a, b)5 marksArithmetic progressions
  • 2026 May/June · Paper 12 · Q10 (a, b)8 marksGeometric progressions · Sum to infinity
  • 2026 May/June · Paper 22 · Q68 marksBinomial expansion

2025

  • 2025 Oct/Nov · Paper 12 · Q3 (a, b)6 marksGeometric progressions · Sum to infinity
  • 2025 Oct/Nov · Paper 22 · Q36 marksBinomial expansion
  • 2025 Oct/Nov · Paper 22 · Q8 (a, b)7 marksArithmetic progressions · Geometric progressions
  • 2025 May/June · Paper 12 · Q7 (a, b)7 marksGeometric progressions · Sum to infinity
  • 2025 May/June · Paper 22 · Q7 (a, b)9 marksArithmetic progressions
  • 2025 May/June · Paper 22 · Q109 marksBinomial expansion

2024

  • 2024 Oct/Nov · Paper 12 · Q6 (a, b)5 marksBinomial expansion
  • 2024 Oct/Nov · Paper 12 · Q10 (a, b)11 marksArithmetic progressions · Sum to infinity
  • 2024 May/June · Paper 12 · Q10 (a(i), a(ii), b(i), b(ii), c)12 marksArithmetic progressions · Geometric progressions · Sum to infinity
  • 2024 May/June · Paper 22 · Q118 marksBinomial expansion

2023

  • 2023 Oct/Nov · Paper 12 · Q4 (a, b)7 marksBinomial expansion
  • 2023 Oct/Nov · Paper 22 · Q9 (a, b)10 marksArithmetic progressions · Geometric progressions
  • 2023 May/June · Paper 12 · Q10 (a(i), a(ii), a(iii), b, c)13 marksArithmetic progressions · Geometric progressions · Sum to infinity
  • 2023 May/June · Paper 22 · Q6 (a(i), a(ii), b)9 marksBinomial expansion

2022

  • 2022 Oct/Nov · Paper 12 · Q10 (a, b, c)8 marksArithmetic progressions
  • 2022 Oct/Nov · Paper 12 · Q129 marksBinomial expansion
  • 2022 Oct/Nov · Paper 22 · Q10 (a, b)8 marksGeometric progressions
  • 2022 May/June · Paper 12 · Q58 marksBinomial expansion
  • 2022 May/June · Paper 22 · Q10 (a(i), a(ii), b)13 marksArithmetic progressions · Geometric progressions

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Binomial expansion · Core · Paper 1 style · no calculator · 3 marks
  1. (a) Find the first three terms in the expansion of (2 + x)⁵ in ascending powers of x. Simplify each coefficient. [3]
Show answer
  1. (a) 32 + 80x + 80x²
    B1 32; B1 ⁵C₁·2⁴x = 80x; B1 ⁵C₂·2³x² = 80x²
Q2Binomial expansion · Core · Paper 2 style · calculator · 6 marks
  1. (a) Find the first three terms in the expansion of (1 − 3x)⁷ in ascending powers of x. [3]
  2. (b) In the expansion of (1 − 2x)ⁿ, where n is a positive integer, the coefficient of x² is 180. Find the value of n. [3]
Show answer
  1. (a) 1 − 21x + 189x²
    B1 1; B1 ⁷C₁(−3x) = −21x; B1 ⁷C₂(9x²) = 189x²
  2. (b) n = 10
    ⁿC₂·4 = 180 M1; n(n − 1) = 90 A1; n = 10 (reject −9) A1
Q3Binomial expansion · Core · Paper 1 style · no calculator · 3 marks
  1. (a) Find the coefficient of x³ in the expansion of (3 − 2x)⁶. [3]
Show answer
  1. (a) −4320
    ⁶C₃ = 20 B1; 20 × 3³ × (−2)³ M1; −4320 A1
Q4Binomial expansion · Core · Paper 1 style · no calculator · 6 marks
  1. (a) Find the first four terms in the expansion of (1 + x/2)⁶ in ascending powers of x. [4]
  2. (b) Hence find the coefficient of x³ in the expansion of (2 − x)(1 + x/2)⁶. [2]
Show answer
  1. (a) 1 + 3x + (15/4)x² + (5/2)x³
    B1 1 + 3x; B1 15/4 x²; B1 5/2 x³; B1 all correct and simplified
  2. (b) 5/4
    2 × 5/2 − 15/4 M1; 5/4 A1
Q5Binomial expansion · Standard · Paper 1 style · no calculator · 4 marks
  1. (a) Find the term independent of x in the expansion of (2x − 1/x²)⁶. [4]
Show answer
  1. (a) 240
    General term ⁶Cᵣ(2x)⁶⁻ʳ(−x⁻²)ʳ M1; power of x: 6 − 3r = 0 so r = 2 M1; ⁶C₂ × 2⁴ × (−1)² M1; 240 A1
Q6Binomial expansion · Standard · Paper 2 style · calculator · 7 marks
  1. (a) The first three terms in the expansion of (1 + ax)ⁿ, where n is a positive integer and a is a constant, are 1 + 12x + 60x². Find the value of n and the value of a. [5]
  2. (b) Find the coefficient of x³ in this expansion. [2]
Show answer
  1. (a) n = 6, a = 2
    na = 12 B1; n(n − 1)a²/2 = 60 B1; so (n − 1)a = 10 M1 (dividing); a = 2 A1; n = 6 A1
  2. (b) 160
    ⁶C₃ × 2³ M1; 160 A1
Q7Binomial expansion · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) Find the first three terms in the expansion of (2 − x)⁵ in ascending powers of x. [3]
  2. (b) Hence find the coefficient of x² in the expansion of (1 + 3x)(2 − x)⁵. [2]
Show answer
  1. (a) 32 − 80x + 80x²
    B1 B1 B1, one for each term
  2. (b) −160
    80 + 3 × (−80) M1; −160 A1
Q8Binomial expansion · Standard · Paper 2 style · calculator · 5 marks
  1. (a) Find the first three terms in the expansion of (1 − x/4)⁸ in ascending powers of x. Simplify each coefficient. [3]
  2. (b) Use your expansion, with a suitable value of x, to estimate 0.99⁸. Give your answer correct to 4 decimal places. [2]
Show answer
  1. (a) 1 − 2x + (7/4)x²
    B1 1; B1 −2x; B1 ⁸C₂/16 = 28/16 = 7/4
  2. (b) 0.9228
    1 − x/4 = 0.99 so x = 0.04 M1; 1 − 0.08 + 0.0028 = 0.9228 A1
Q9Binomial expansion · Challenge · Paper 2 style · calculator · 6 marks
  1. (a) In the expansion of (3 + kx)⁵, where k is a positive constant, the coefficient of x² is twice the coefficient of x. Find the value of k. [4]
  2. (b) Using your value of k, find the coefficient of x³. [2]
Show answer
  1. (a) k = 3
    coefficient of x: ⁵C₁ × 3⁴k = 405k B1; coefficient of x²: ⁵C₂ × 3³k² = 270k² B1; 270k² = 810k M1; k = 3 (k ≠ 0) A1
  2. (b) 2430
    ⁵C₃ × 3² × 3³ M1; 2430 A1
Q10Binomial expansion · Challenge · Paper 1 style · no calculator · 6 marks
  1. (a) Find the term independent of x in the expansion of (x² + 2/x)⁶ (1 − x³). [6]
Show answer
  1. (a) 48
    General term ⁶Cᵣ 2ʳ x¹²⁻³ʳ M1; constant term of (x² + 2/x)⁶: r = 4, 15 × 16 = 240 A1; x⁻³ term: r = 5, 6 × 32 = 192, i.e. 192x⁻³ A1; constant in product = 1 × 240 + (−x³)(192x⁻³) M1; 240 − 192 = 48 A1 (allow B1 for identifying both terms needed)
Q11Binomial expansion · Challenge · Paper 2 style · calculator · 6 marks
  1. (a) In the expansion of (2 + x)ⁿ, where n ≥ 3 is an integer, the coefficients of x² and x³ are equal. Find the value of n. [4]
  2. (b) Find this common coefficient. [2]
Show answer
  1. (a) n = 8
    ⁿC₂ 2ⁿ⁻² = ⁿC₃ 2ⁿ⁻³ M1; 2 × n(n − 1)/2 = n(n − 1)(n − 2)/6 M1 (cancelling); n − 2 = 6 A1; n = 8 A1
  2. (b) 1792
    ⁸C₂ × 2⁶ = 28 × 64 M1; 1792 A1
Q12Arithmetic progressions · Core · Paper 1 style · no calculator · 4 marks
  1. (a) An arithmetic progression has first term 7 and common difference 4. Find the 20th term. [2]
  2. (b) Find the sum of the first 20 terms. [2]
Show answer
  1. (a) 83
    7 + 19 × 4 M1; 83 A1
  2. (b) 900
    20/2 × (7 + 83) M1; 900 A1
Q13Arithmetic progressions · Core · Paper 2 style · calculator · 4 marks
  1. (a) Ayesha saves Rs 2000 in the first month. Each month after that she saves Rs 250 more than in the month before. Find the total amount she saves in the first 24 months. [2]
  2. (b) Find the number of the first month in which she saves more than Rs 6000. [2]
Show answer
  1. (a) Rs 117 000
    24/2 × (2 × 2000 + 23 × 250) M1; 117 000 A1
  2. (b) the 18th month
    2000 + 250(n − 1) > 6000 M1; n − 1 > 16 so n = 18 A1 (month 17 is exactly 6000)
Q14Arithmetic progressions · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) The 4th term of an arithmetic progression is 17 and the 11th term is 52. Find the first term and the common difference. [3]
  2. (b) Find the sum of the first 16 terms. [2]
Show answer
  1. (a) a = 2, d = 5
    a + 3d = 17 and a + 10d = 52 M1; d = 5 A1; a = 2 A1
  2. (b) 632
    8 × (2 × 2 + 15 × 5) M1; 632 A1
Q15Arithmetic progressions · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) Show that lg 2, lg 8, lg 32, … are the first three terms of an arithmetic progression and state its common difference in the form k lg 2. [2]
  2. (b) Find the sum of the first 20 terms, giving your answer in the form p lg 2, where p is an integer. [3]
Show answer
  1. (a) terms are lg 2, 3 lg 2, 5 lg 2; common difference 2 lg 2
    lg 8 = 3 lg 2, lg 32 = 5 lg 2 M1; equal differences 2 lg 2 A1
  2. (b) 400 lg 2
    S = 20/2 × (2 lg 2 + 19 × 2 lg 2) M1 A1; 400 lg 2 A1
Q16Arithmetic progressions · Standard · Paper 2 style · calculator · 4 marks
  1. (a) Find the sum of all the integers between 100 and 400 that are divisible by 7. [4]
Show answer
  1. (a) 10 836
    first 105, last 399 B1; n = (399 − 105)/7 + 1 = 43 M1 A1; 43/2 × (105 + 399) = 10 836 A1
Q17Arithmetic progressions · Challenge · Paper 2 style · calculator · 6 marks
  1. (a) The sum of the first n terms of a progression is given by Sₙ = 3n² + 5n. Find the first term and the second term, and show that the progression is arithmetic. [4]
  2. (b) Find the 15th term. [2]
Show answer
  1. (a) u₁ = 8, u₂ = 14; uₙ = Sₙ − Sₙ₋₁ = 6n + 2, so the difference is always 6
    S₁ = 8 B1; S₂ = 22 so u₂ = 14 B1; Sₙ − Sₙ₋₁ = 6n + 2 M1; linear in n, so common difference 6 A1
  2. (b) 92
    8 + 14 × 6 or 6(15) + 2 M1; 92 A1
Q18Arithmetic progressions · Challenge · Paper 2 style · calculator · 5 marks
  1. (a) A cricket stadium has 40 seats in its front row. Each row behind has 3 more seats than the row in front of it. Find the least number of rows needed so that the stadium holds at least 2500 people. [5]
Show answer
  1. (a) 30 rows
    Sₙ = n/2(80 + 3(n − 1)) ≥ 2500 M1; 3n² + 77n − 5000 ≥ 0 A1; n ≥ 29.96… M1 A1; n = 30 A1 (S₂₉ = 2378, S₃₀ = 2505)
Q19Arithmetic progressions · Challenge · Paper 1 style · no calculator · 6 marks
  1. (a) The first three terms of an arithmetic progression are 2k + 1, 5k − 4 and 6k + 3. Find the value of k. [3]
  2. (b) Find the sum of the first 10 terms. [3]
Show answer
  1. (a) k = 6
    (5k − 4) − (2k + 1) = (6k + 3) − (5k − 4) M1; 3k − 5 = k + 7 A1; k = 6 A1
  2. (b) 715
    terms 13, 26, 39 so a = 13, d = 13 B1; 5 × (26 + 9 × 13) M1; 715 A1
Q20Geometric progressions · Core · Paper 2 style · calculator · 4 marks
  1. (a) A geometric progression has first term 5 and common ratio 3. Find the 8th term. [2]
  2. (b) Find the sum of the first 8 terms. [2]
Show answer
  1. (a) 10 935
    5 × 3⁷ M1; 10 935 A1
  2. (b) 16 400
    5(3⁸ − 1)/(3 − 1) M1; 16 400 A1
Q21Geometric progressions · Core · Paper 1 style · no calculator · 5 marks
  1. (a) The 2nd term of a geometric progression is 6 and the 5th term is 162. Find the first term and the common ratio. [3]
  2. (b) Find the sum of the first 6 terms. [2]
Show answer
  1. (a) a = 2, r = 3
    ar = 6, ar⁴ = 162 M1; r³ = 27, r = 3 A1; a = 2 A1
  2. (b) 728
    2(3⁶ − 1)/2 M1; 728 A1
Q22Geometric progressions · Standard · Paper 2 style · calculator · 5 marks
  1. (a) The population of a town in Punjab is 85 000 and is expected to grow by 3% each year. Find the expected population after 10 years. Give your answer correct to 3 significant figures. [2]
  2. (b) Find the least whole number of years after which the population is first expected to be more than 150 000. [3]
Show answer
  1. (a) 114 000
    85 000 × 1.03¹⁰ M1; 114 233 → 114 000 A1
  2. (b) 20 years
    85 000 × 1.03ⁿ > 150 000 M1; n > lg(150/85)/lg 1.03 = 19.2 M1; n = 20 A1
Q23Geometric progressions · Standard · Paper 2 style · calculator · 5 marks
  1. (a) An engineer in Islamabad starts on a salary of Rs 900 000 per year. Her salary increases by 5% each year. Find the total salary she earns in her first 12 years. Give your answer correct to the nearest Rs 1000. [3]
  2. (b) Find her salary in the 12th year, correct to the nearest Rs 1000. [2]
Show answer
  1. (a) Rs 14 325 000
    GP a = 900 000, r = 1.05 B1; 900 000(1.05¹² − 1)/0.05 M1; 14 325 414 → Rs 14 325 000 A1
  2. (b) Rs 1 539 000
    900 000 × 1.05¹¹ M1; 1 539 305 → Rs 1 539 000 A1
Q24Geometric progressions · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) A geometric progression has first term 3. The sum of its first three terms is 21. Find the two possible values of the common ratio. [3]
  2. (b) Find the two possible values of the 5th term. [2]
Show answer
  1. (a) r = 2 or r = −3
    3 + 3r + 3r² = 21 M1; r² + r − 6 = 0 A1; (r − 2)(r + 3) = 0, r = 2 or −3 A1
  2. (b) 48 or 243
    3 × 2⁴ = 48 B1; 3 × (−3)⁴ = 243 B1
Q25Geometric progressions · Challenge · Paper 2 style · calculator · 4 marks
  1. (a) Find the least number of terms of the geometric progression 3 + 6 + 12 + … whose sum is greater than 1 000 000. [4]
Show answer
  1. (a) 19 terms
    3(2ⁿ − 1) > 1 000 000 M1; 2ⁿ > 333 334.3 A1; n > lg 333 334.3 / lg 2 = 18.35 M1; n = 19 A1
Q26Geometric progressions · Challenge · Paper 1 style · no calculator · 6 marks
  1. (a) The first three terms of a geometric progression are x + 6, x and x − 4, where x > 4. Find the value of x. [3]
  2. (b) Find the common ratio and the 6th term, giving the 6th term as a fraction in its simplest form. [3]
Show answer
  1. (a) x = 12
    x² = (x + 6)(x − 4) M1; x² = x² + 2x − 24 A1; x = 12 A1
  2. (b) r = 2/3; 6th term = 64/27
    terms 18, 12, 8 so r = 2/3 B1; 18 × (2/3)⁵ M1; 64/27 A1
Q27Sum to infinity · Core · Paper 1 style · no calculator · 2 marks
  1. (a) Find the sum to infinity of the geometric progression 24, −12, 6, … [2]
Show answer
  1. (a) 16
    r = −1/2 B1; 24/(1 + 1/2) = 16 B1
Q28Sum to infinity · Standard · Paper 2 style · calculator · 4 marks
  1. (a) A ball is dropped from a height of 8 m onto level ground. After each bounce it rises to 3/5 of the height from which it last fell. Find the total distance travelled by the ball before it comes to rest, assuming it bounces forever. [4]
Show answer
  1. (a) 32 m
    first rise 4.8 m B1; distance = 8 + 2 × (4.8 + 2.88 + …) M1; 2 × 4.8/(1 − 0.6) = 24 M1; total 32 m A1
Q29Sum to infinity · Standard · Paper 1 style · no calculator · 5 marks
  1. (a) The second term of a geometric progression is 6 and its sum to infinity is 32. Find the two possible values of the common ratio and the corresponding first terms. [5]
Show answer
  1. (a) r = 1/4, a = 24 or r = 3/4, a = 8
    ar = 6 and a/(1 − r) = 32 B1; 6/r = 32(1 − r) M1; 16r² − 16r + 3 = 0 A1; (4r − 1)(4r − 3) = 0 M1; r = 1/4, a = 24; r = 3/4, a = 8 A1
Q30Sum to infinity · Challenge · Paper 1 style · no calculator · 8 marks
  1. (a) A geometric progression has sum to infinity 27. The sum of its first three terms is 19. Find the common ratio and the first term. [5]
  2. (b) Find the least value of n for which the sum of the first n terms is more than 26. [3]
Show answer
  1. (a) r = 2/3, a = 9
    a/(1 − r) = 27 B1; a(1 − r³)/(1 − r) = 19 B1; dividing, 1 − r³ = 19/27 M1; r³ = 8/27 so r = 2/3 A1; a = 27 × 1/3 = 9 A1
  2. (b) n = 9
    27(1 − (2/3)ⁿ) > 26 M1; (2/3)ⁿ < 1/27 M1; n > 8.13 so n = 9 A1

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Binomial expansion Banker9 questions · 9/9 sessions
Arithmetic progressions Banker9 questions · 9/9 sessions
Geometric progressions Banker9 questions · 8/9 sessions
Sum to infinity 6 questions · 6/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Binomial expansion Banker699/9
Arithmetic progressions Banker549/9
Geometric progressions Banker428/9
Sum to infinity 216/9

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