0606/12 Feb/Mar 2026, Question 4: Arithmetic Progressions in Problems
Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 3 marks
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The question, in short
An arithmetic progression has common difference −2 and equal sums for the first 4 and first 10 terms. Find the first term.
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Solution · 3 marks
- 1S₄ = (4/2)(2a + 3(−2)) = 4a − 12
- 2S₁₀ = (10/2)(2a + 9(−2)) = 10a − 90
- 34a − 12 = 10a − 90
- 46a = 78, a = 13
Answer13
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