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4PM1/01R Jun 2023, Question 8: Geometric Progressions and Sum to Infinity

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 10 marks

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The question, in short

Geometric series with nth term Uₙ = (25/4)(3/5)ⁿ: calculate a specific term, rewrite the partial sum in a required form, and find the smallest n meeting an accuracy condition.

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

Part (a) · 1 mark

Calculate the exact value of the fifth term of the series.

  1. 1U₅ = (25/4) × (3/5)⁵ = (25/4) × 243/3125
  2. 2= 243/500

Answer243/500

Part (b) · 3 marks

Express the sum of the first n terms as a summation with integer constants A and B in the required form.

  1. 1U₁ = (25/4)(3/5) = 15/4
  2. 2Common ratio r = 3/5
  3. 3Uᵣ = (15/4)(3/5)^(r − 1)
  4. 4Sₙ = Σ from r = 1 to n of (15/4)(3/5)^(r − 1): A = 15, B = 4

AnswerSₙ = Σ from r = 1 to n of (15/4)(3/5)^(r − 1), so A = 15, B = 4 — shown

Part (c) · 6 marks

Determine the smallest number of terms needed so that the difference between the infinite sum and the partial sum is below 0.045.

  1. 1S = (15/4)/(1 − 3/5) = 75/8 = 9.375
  2. 2Sₙ = (15/4)(1 − (3/5)ⁿ)/(1 − 3/5) = 9.375(1 − (3/5)ⁿ)
  3. 3S − Sₙ = 9.375(3/5)ⁿ < 0.045, so (3/5)ⁿ < 3/625
  4. 4n lg(3/5) < lg(3/625); lg(3/5) < 0 so n > 10.45
  5. 5Least n = 11

Answern = 11

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