0606/11 May/Jun 2025, Question 11: Arithmetic Progressions in Problems
Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 9 marks
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The question, in short
Arithmetic progression with common difference d has 3rd term 10. Find expressions for first two terms, then use squared first three terms summing to 140 to find sum of first 200 terms for the smaller d.
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Part (a) · 2 marks
Write expressions for the 1st and 2nd terms in terms of d only, given the 3rd term is 10.
- 13rd term = a + 2d = 10, so a = 10 − 2d
- 22nd term = a + d = 10 − d
Answer1st term 10 − 2d, 2nd term 10 − d
Part (b) · 7 marks
Given the first three terms squared sum to 140, find the two possible d values and the sum of the first 200 terms using the smaller d.
- 1(10 − 2d)² + (10 − d)² + 10² = 140
- 2100 − 40d + 4d² + 100 − 20d + d² + 100 = 140
- 35d² − 60d + 160 = 0, so d² − 12d + 32 = 0
- 4(d − 4)(d − 8) = 0, d = 4 or 8; smaller d = 4
- 5a = 10 − 2(4) = 2
- 6S₂₀₀ = (200/2){2(2) + 199 × 4} = 100 × 800 = 80 000
Answer80 000
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