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4PM1/02R Jan 2023, Question 9: Geometric Progressions and Sum to Infinity

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 12 marks

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The question, in short

Geometric series with r > 0 has third term 27/2 and first-three-term sum 57/2; work with Sₙ to prove a summation form, a k inequality, and find least k.

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

Part (a) · 8 marks

Show that the sum to n terms equals the sum from j = 1 to n of 4(3/2)^j.

  1. 1ar² = 27/2
  2. 2a + ar + 27/2 = 57/2 → a(1 + r) = 15 → a = 15/(1 + r)
  3. 3(15/(1 + r))r² = 27/2 → 10r² − 9r − 9 = 0
  4. 4(5r + 3)(2r − 3) = 0, r > 0 so r = 3/2
  5. 5a = 15/(5/2) = 6
  6. 6Term j is 6(3/2)^(j − 1) = 4(3/2)^j, so Sₙ = Σ from j = 1 to n of 4(3/2)^j

AnswerSₙ = Σ from j = 1 to n of 4(3/2)^j shown (a = 6, r = 3/2)

Part (b) · 3 marks

Given Sₖ > 50 000, show the least k satisfies k > lg(12 503/3)/lg(3/2).

  1. 16(1.5^k − 1)/(1.5 − 1) > 50 000
  2. 212(1.5^k − 1) > 50 000 → 1.5^k > 12 503/3
  3. 3k lg 1.5 > lg(12 503/3)
  4. 4k > lg(12 503/3)/lg(3/2)

Answerk > lg(12 503/3)/lg(3/2) shown

Part (c) · 1 mark

Use the inequality to find the least integer value of k.

  1. 1k > 20.556…
  2. 2Least k = 21

Answerk = 21

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