Lemmara logoLemmaraWorked solution

4PM1/02 Nov 2021, Question 7: Geometric Progressions and Sum to Infinity

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 10 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Geometric series with first term a and ratio r: first three terms sum to 61/6 and sum to infinity is 125/6; show r = 4/5, find a, then use Sₙ > 19.8 to find least n.

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

Part (a) · 6 marks

(i) Show that r = 4/5 using the two given sums. (ii) Find the value of a.

  1. 1(i) a(1 − r³)/(1 − r) = 61/6 and a/(1 − r) = 125/6
  2. 2Divide: 1 − r³ = (61/6)/(125/6) = 61/125
  3. 3r³ = 64/125
  4. 4r = 4/5
  5. 5(ii) a/(1 − 4/5) = 125/6
  6. 6a = (125/6)(1/5) = 25/6

Answer(i) r = 4/5 shown; (ii) a = 25/6

Part (b) · 2 marks

Given Sₙ > 19.8, show that n lg(4/5) < lg(31/625).

  1. 1Sₙ = (25/6)(1 − 0.8ⁿ)/(1 − 0.8) = (125/6)(1 − 0.8ⁿ) > 19.8
  2. 21 − 0.8ⁿ > 594/625
  3. 30.8ⁿ < 31/625
  4. 4Take lg: n lg(0.8) < lg(31/625)

Answern lg(4/5) < lg(31/625) shown

Part (c) · 2 marks

Hence find the least value of n.

  1. 1lg(0.8) < 0, so dividing reverses the inequality
  2. 2n > lg(31/625)/lg(0.8) = 13.46…
  3. 3Least n = 14

Answern = 14

Stuck on a step? Read the free lesson: Geometric Progressions and Sum to Infinity

Practise more geometric progressions and sum to infinity questions

Real 4PM1 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on geometric progressions and sum to infinity

Also searched as: 4PM1 2 November 2021 · 4PM1/02 · 4pm1 nov21 2 · question 7.
Lemmara is an independent study tool, not affiliated with or endorsed by Pearson Edexcel. The worked solutions are Lemmara’s own.