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4PM1/01 Jan 2020, Question 1: Arithmetic Progressions in Problems

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 9 marks

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The question, in short

Arithmetic series with nth term tₙ and common difference d satisfies t₂ + t₉ = 0 and t₄ + t₆ + t₁₀ = 14. Work with these conditions, then handle a second series with first term 24 and difference 6.

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

Part (a) · 4 marks

Using the two given equations, prove d = 4 and then determine the series' first term.

  1. 1t₂ + t₉ = 0: a + d + a + 8d = 0, so 2a + 9d = 0
  2. 2t₄ + t₆ + t₁₀ = 14: a + 3d + a + 5d + a + 9d = 14, so 3a + 17d = 14
  3. 3Solve simultaneously: from the first, a = −9d/2
  4. 43(−9d/2) + 17d = 14 gives 7d/2 = 14, so d = 4
  5. 5a = −9 × 4/2 = −18

Answerd = 4 shown; first term a = −18

Part (b) · 5 marks

For the series starting at 24 with difference 6, the first 2n terms total three times the first n terms; determine n.

  1. 1S₂ₙ = (2n/2)[48 + 6(2n − 1)] and Sₙ = (n/2)[48 + 6(n − 1)]
  2. 2n(42 + 12n) = 3 × (n/2)(42 + 6n)
  3. 342n + 12n² = 63n + 9n², so 3n² − 21n = 0
  4. 43n(n − 7) = 0
  5. 5n = 7 (n = 0 is not valid)

Answern = 7

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