4PM1/01R Jun 2025, Question 2: Arithmetic Progressions in Problems
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 7 marks
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The question, in short
An arithmetic series has third term 8 and fifth term 20. Given the sum to n terms exceeds 220, find the smallest possible value of n.
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Solution · 7 marks
- 1a + 2d = 8 and a + 4d = 20
- 22d = 12, so d = 6 and a = −4
- 3Sₙ = (n/2)(2(−4) + (n − 1)6) > 220
- 43n² − 7n − 220 > 0
- 5n = (7 ± √(49 + 2640))/6 gives n = 9.809… or −7.47…
- 6Least integer n = 10 (check: S₉ = 180, S₁₀ = 230)
Answern = 10
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