4PM1/01R Jan 2023, Question 3: Arithmetic Progressions in Problems
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 4 marks
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The question, in short
An arithmetic series starts at 16 with common difference −5; its sum to n terms is Sₙ. Given Sₙ < −450, find the smallest n satisfying this.
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Solution · 4 marks
- 1Sₙ = (n/2)[2 × 16 + (n − 1)(−5)] < −450
- 237n − 5n² < −900
- 35n² − 37n − 900 > 0
- 4n = (37 ± √(37² + 4 × 5 × 900))/10 → n = 17.617… (or −10.2…)
- 5Least integer n = 18
Answern = 18
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