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4PM1/01 Nov 2021, Question 9: Arithmetic Progressions in Problems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 11 marks

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The question, in short

Prove the given summation formula for 5r − 1, use it to evaluate a sum from r = 10 to 20, then solve for n in an equation linking Sₙ and u_(n+1).

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

Part (a) · 3 marks

Show that the sum from r = 1 to n of (5r − 1) equals (n/2)(3 + 5n).

  1. 1First term a = 5(1) − 1 = 4, common difference d = 5
  2. 2Sₙ = (n/2)(2(4) + (n − 1)5)
  3. 3= (n/2)(8 + 5n − 5) = (n/2)(3 + 5n)

AnswerΣ (5r − 1) = (n/2)(3 + 5n) shown

Part (b) · 3 marks

Using part a or otherwise, evaluate the sum of (5r − 1) from r = 10 to 20.

  1. 1Σ from 10 to 20 = Σ from 1 to 20 − Σ from 1 to 9
  2. 2= (20/2)(3 + 100) − (9/2)(3 + 45)
  3. 3= 1030 − 216 = 814

Answer814

Part (c) · 5 marks

Given Sₙ = 12u_(n+1) + 52, find the value of n.

  1. 1u_(n+1) = 5(n + 1) − 1 = 5n + 4
  2. 2(n/2)(3 + 5n) = 12(5n + 4) + 52
  3. 35n² − 117n − 200 = 0
  4. 4(n − 25)(5n + 8) = 0
  5. 5n = 25 (n = −8/5 rejected)

Answern = 25

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