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WMA12/01 Jun 2023, Question 6: Geometric Progressions and Sum to Infinity

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 9 marks

The question, in short

Given three monthly download counts in thousands, 2k − 15, k, k + 4, modelled as a geometric series, find k, the 4th month downloads, and when cumulative downloads first exceed 3 million.

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

Part (a) · 2 marks

Show that the geometric condition on the three terms leads to k² − 7k − 60 = 0.

  1. 1Common ratio: k/(2k − 15) = (k + 4)/k
  2. 2k² = (k + 4)(2k − 15)
  3. 3k² = 2k² − 7k − 60
  4. 4k² − 7k − 60 = 0

AnswerShown: k/(2k − 15) = (k + 4)/k gives k² = 2k² − 7k − 60, so k² − 7k − 60 = 0

Part (b) · 4 marks

Predict the number of downloads, in thousands, in the 4th month.

  1. 1(k − 12)(k + 5) = 0; terms must be positive, so k = 12
  2. 2Terms are 9, 12, 16 (thousand)
  3. 3r = 12/9 = 4/3
  4. 44th term = 16 × 4/3 = 21.333... thousand
  5. 5About 21 300 downloads

Answer21 300 downloads (21 333 to the nearest whole number)

Part (c) · 3 marks

Calculate the month N in which total downloads first exceed 3 million.

  1. 1a = 9, r = 4/3, and 3 million = 3000 thousand
  2. 2S_N = 9((4/3)ᴺ − 1)/(4/3 − 1) = 27((4/3)ᴺ − 1) > 3000
  3. 3(4/3)ᴺ > 1 + 3000/27 = 1009/9
  4. 4N > log(1009/9) / log(4/3) = 16.405...
  5. 5So N = 17

AnswerN = 17

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Also searched as: WMA12 June 2023 · WMA12/01 · wma12 jun23 · IAS Maths Pure Mathematics 2 June 2023 · question 6.
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