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4PM1/01R Jun 2019, Question 6: Arithmetic Progressions in Problems

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

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The question, in short

Prove the sum of 4r − 3 from r = 1 to n equals n(2n − 1), then use it to find the smallest n where the sum exceeds 1000, and solve 18 + 3t₍ₙ₊₇₎ = S₍ₙ₊₄₎ for n.

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 (4r − 3) equals n(2n − 1).

  1. 1First term a = 4(1) − 3 = 1, d = 4
  2. 2Sum = (n/2)[2(1) + (n − 1)4]
  3. 3= (n/2)(4n − 2) = n(2n − 1)

AnswerΣ(4r − 3) = n(2n − 1) shown

Part (b) · 3 marks

Using the result, or otherwise, find the least n for which the sum from r = 1 to n of (4r − 3) is greater than 1000.

  1. 1n(2n − 1) > 1000 → 2n² − n − 1000 > 0
  2. 2n = (1 + √(1 + 8000))/4 = 22.612…
  3. 3Least n = 23

Answern = 23

Part (c) · 4 marks

Given Sₙ = n(2n − 1) and tₙ = 4n − 3, solve 18 + 3t₍ₙ₊₇₎ = S₍ₙ₊₄₎ to find n.

  1. 118 + 3[4(n + 7) − 3] = (n + 4)[2(n + 4) − 1]
  2. 218 + 12n + 75 = 2n² + 15n + 28
  3. 32n² + 3n − 65 = 0
  4. 4(2n + 13)(n − 5) = 0, so n = 5 (reject −13/2)

Answern = 5

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