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0606/12 May/Jun 2025, Question 7: Geometric Progressions and Sum to Infinity

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 7 marks

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The question, in short

A GP has 4th term 8k⁶/27 and 6th term 32k¹⁰/243, with positive common ratio. Find the ratio and first term, then use a sum to infinity of 3 to find k.

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

Part (a) · 4 marks

Work out the positive common ratio expressed using k, and also determine the first term of the progression.

  1. 1ar³ = 8k⁶/27 and ar⁵ = 32k¹⁰/243
  2. 2r² = (32k¹⁰/243) ÷ (8k⁶/27) = 4k⁴/9
  3. 3r > 0, so r = 2k²/3
  4. 4a × 8k⁶/27 = 8k⁶/27, so a = 1

Answerr = 2k²/3, a = 1

Part (b) · 3 marks

Using the given sum to infinity equal to 3, determine all possible values of k.

  1. 1S∞ = a/(1 − r): 1/(1 − 2k²/3) = 3
  2. 21 − 2k²/3 = 1/3
  3. 3k² = 1
  4. 4k = ±1

Answerk = ±1

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Also searched as: 0606/12/M/J/25 · 0606_s25_qp_12 · 0606 s25 12 · question 7.
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