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WMA12/01 Oct 2021, Question 7: Geometric Progressions and Sum to Infinity

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 10 marks

The question, in short

For a geometric sequence starting at 4 with ratio 6, find the smallest n where the nth term exceeds 10¹⁰⁰; then for another geometric sequence with second term −6 and infinite sum 25, derive and solve a quadratic in r, choose r, and find the first four terms' sum.

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

Part (i) · 3 marks

Find the least n for which the nth term of the sequence with first term 4 and common ratio 6 is larger than 10¹⁰⁰.

  1. 14 × 6ⁿ⁻¹ > 10¹⁰⁰
  2. 2log 4 + (n − 1) log 6 > 100
  3. 3n − 1 > (100 − log 4)/log 6 = 127.74…
  4. 4n − 1 = 128, so n = 129

Answern = 129

Part (iia) · 3 marks

Using the second term −6 and infinite sum 25, derive the quadratic equation 25r² − 25r − 6 = 0.

  1. 1ar = −6 and a/(1 − r) = 25
  2. 2a = −6/r, so −6/(r(1 − r)) = 25
  3. 3−6 = 25r − 25r²
  4. 425r² − 25r − 6 = 0

AnswerShown: −6/(r(1 − r)) = 25 ⇒ 25r² − 25r − 6 = 0

Part (iib) · 1 mark

Give the two roots of 25r² − 25r − 6 = 0.

  1. 1(5r − 6)(5r + 1) = 0
  2. 2r = 6/5 or r = −1/5

Answerr = 6/5, r = −1/5

Part (iic) · 1 mark

Decide which root is the valid common ratio r, justifying why the other is rejected.

  1. 1A sum to infinity only exists when |r| < 1
  2. 26/5 > 1, so r = −1/5

Answerr = −1/5, because |r| < 1 is needed for the sum to infinity to exist

Part (iid) · 2 marks

Calculate the total of the first 4 terms of this series using the chosen r.

  1. 1a = −6 ÷ (−1/5) = 30
  2. 2S₄ = 30(1 − (−1/5)⁴)/(1 + 1/5)
  3. 3= 30 × (624/625) ÷ (6/5)
  4. 4= 624/25 = 24.96

Answer24.96 (= 624/25)

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Also searched as: WMA12 October 2021 · WMA12/01 · wma12 oct21 · IAS Maths Pure Mathematics 2 October 2021 · question 7.
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