WMA12/01 Jun 2025, Question 7: Arithmetic Series: Proofs and Conditions

Part (ia) · 2 marks

Find the sum to infinity of the geometric series starting 10 + 8 + 6.4 + …

1. 1a = 10, r = 8/10 = 0.8
2. 2S∞ = a/(1 − r) = 10/0.2 = 50

Answer50

Part (ib) · 3 marks

Given the kth term is below 0.0005, use algebra to find the smallest possible k.

1. 1kth term = 10 × 0.8ᵏ⁻¹ < 0.0005, so 0.8ᵏ⁻¹ < 0.00005
2. 2(k − 1) log 0.8 < log 0.00005; log 0.8 is negative, so the inequality flips
3. 3k − 1 > log 0.00005 ÷ log 0.8 = 44.38…
4. 4k > 45.38…, so the smallest k is 46

Answerk = 46

Part (ii) · 4 marks

For the arithmetic series 850 + 843 + 836 + …, find the greatest possible value of Sₙ.

1. 1a = 850, d = −7; the sum is greatest when we stop at the last positive term
2. 2850 + (n − 1)(−7) = 0 gives n = 122.4…
3. 3So the last positive term is term 122 (it equals 3)
4. 4S₁₂₂ = (122/2) × {2 × 850 + 121 × (−7)} = 61 × 853 = 52033

Answer52033

More past-paper questions on arithmetic series: proofs and conditions

• WMA12/01 Jan 2023 Q3 · 7 marks
• WMA12/01 Oct 2021 Q5 · 6 marks
• WMA12/01 Jan 2020 Q8 · 7 marks
• WMA12/01 Oct 2020 Q5 · 11 marks