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4PM1/02 Jun 2019, Question 7: Geometric Progressions and Sum to Infinity

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 9 marks

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The question, in short

For an arithmetic series with nth partial sum Aₙ = Σ(4r + 5) and a geometric series with Gₙ = Σ4(3)^(r−1), find initial terms, common difference/ratio, and solve A₁₄ − 6 = Gₙ for n.

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Part (a) · 2 marks

For the arithmetic series summing 4r + 5, state its first term and its common difference.

  1. 1r = 1: first term = 4 + 5 = 9
  2. 2Each term is 4 more than the last: d = 4

Answer(i) 9 (ii) 4

Part (b) · 2 marks

For the geometric series summing 4(3)^(r−1), state its first term and its common ratio.

  1. 1r = 1: first term = 4(3)⁰ = 4
  2. 2Common ratio = 3

Answer(i) 4 (ii) 3

Part (c) · 5 marks

Find the value of n satisfying A₁₄ − 6 = Gₙ.

  1. 1A₁₄ = (14/2)(2 × 9 + 13 × 4) = 490
  2. 2490 − 6 = 484 = 4(3ⁿ − 1)/(3 − 1)
  3. 33ⁿ − 1 = 242 → 3ⁿ = 243
  4. 4n = 5

Answern = 5

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