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4PM1/01R Jun 2024, Question 5: Arithmetic Progressions in Problems

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

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The question, in short

Arithmetic series A has first 10 terms summing to 36k + 1 and 6th term 4k + 1; find its common difference and first term, then k, then solve an equation linking Sₙ and U₍ₙ₊₁₀₎.

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

Part (a) · 5 marks

Find the common difference in terms of k, then show the first term equals −8.

  1. 16th term: a + 5d = 4k + 1
  2. 2Sum of 10 terms: 5(2a + 9d) = 36k + 1, so 10a + 45d = 36k + 1
  3. 310 × first equation: 10a + 50d = 40k + 10
  4. 4Subtract: 5d = 4k + 9, so d = (4k + 9)/5
  5. 5a = 4k + 1 − (4k + 9) = −8

Answer(i) d = (4k + 9)/5 (ii) a = −8 shown

Part (b) · 2 marks

Given the 4th term is 7, show that k = 4.

  1. 14th term: −8 + 3d = 7, so d = 5
  2. 2(4k + 9)/5 = 5 gives 4k + 9 = 25
  3. 3k = 4

Answerk = 4 shown

Part (c) · 4 marks

Find n satisfying Sₙ = 5U₍ₙ₊₁₀₎ + 105.

  1. 1d = (4 × 4 + 9)/5 = 5
  2. 2(n/2)(2(−8) + 5(n − 1)) = 5[−8 + 5(n + 9)] + 105
  3. 3(5n² − 21n)/2 = 25n + 290
  4. 45n² − 71n − 580 = 0
  5. 5(5n + 29)(n − 20) = 0, so n = 20

Answern = 20

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