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0606/12 May/Jun 2026, Question 4: Arithmetic Progressions in Problems

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 5 marks

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The question, in short

For an arithmetic sequence with first term 400 and common difference −2, work out two later terms and the total of the terms between them.

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Part (a) · 2 marks

Work out the values of the 81st and 100th terms of the progression.

  1. 1u₈₁ = 400 + 80(−2) = 240
  2. 2u₁₀₀ = 400 + 99(−2) = 202

Answeru₈₁ = 240, u₁₀₀ = 202

Part (b) · 3 marks

Work out the sum of the consecutive terms starting at the 81st and ending at the 100th.

  1. 1There are 100 − 81 + 1 = 20 terms
  2. 2Sum = (20/2)(240 + 202)
  3. 3= 10 × 442 = 4420

Answer4420

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Also searched as: 0606/12/M/J/26 · 0606_s26_qp_12 · 0606 s26 12 · question 4.
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