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0606/23 May/Jun 2025, Question 9: Geometric Progressions and Sum to Infinity

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 12 marks

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The question, in short

An arithmetic progression starts at 9, ends at 159 and totals 2604; its 12th and 8th terms seed a geometric progression. Find the sum of that geometric progression's first 6 terms.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 7 marks

Using the given first term, last term and total of the arithmetic progression, work out the geometric progression's first term and ratio, then the sum of its first 6 terms.

  1. 1(n/2)(9 + 159) = 2604, so n = 31
  2. 2159 = 9 + 30d, so d = 5
  3. 3u₁₂ = 9 + 11 × 5 = 64 and u₈ = 9 + 7 × 5 = 44
  4. 4r = 44/64 = 11/16
  5. 5S₆ = 64(1 − (11/16)⁶)/(1 − 11/16) ≈ 183

Answer183 (183.2)

Part (bi) · 1 mark

For a geometric progression with first term sin θ and ratio cos θ, 45° ≤ θ ≤ 135°, show its infinite sum exists.

  1. 1cos 45° = 0.707… and cos 135° = −0.707…
  2. 2So −0.707 ≤ cos θ ≤ 0.707, i.e. −1 < r < 1
  3. 3|r| < 1, so the sum to infinity exists

Answer−1 < cos θ < 1 for 45° ≤ θ ≤ 135°, so |r| < 1 and the sum to infinity exists

Part (bii) · 4 marks

Show that the infinite sum of this geometric progression equals cosec θ + cot θ.

  1. 1S∞ = a/(1 − r) = sin θ/(1 − cos θ)
  2. 2Multiply by (1 + cos θ)/(1 + cos θ): sin θ(1 + cos θ)/(1 − cos²θ)
  3. 3= sin θ(1 + cos θ)/sin²θ = (1 + cos θ)/sin θ
  4. 4= 1/sin θ + cos θ/sin θ = cosec θ + cot θ

AnswerS∞ = cosec θ + cot θ shown

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Also searched as: 0606/23/M/J/25 · 0606_s25_qp_23 · 0606 s25 23 · question 9.
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