Chapter: Arithmetic Progression· Mathematics Part 1 (Algebra)

Asked in: March 2025 board paperA farmer borrows ₹ 1,000 and agrees to repay with a total interest of ₹ 140, in 12 instalments. Each instalment being less than the preceding instalment by ₹ 10. What should be the amount of his first and last instalme…

Asked in: March 2025 board paper

A farmer borrows ₹ 1,000 and agrees to repay with a total interest of ₹ 140, in 12 instalments. Each instalment being less than the preceding instalment by ₹ 10. What should be the amount of his first and last instalment ?

Doubtly Asked 29 Oct 2026 Practice test ssc 10th Save WhatsApp Report

1 answer

Best answer

Answer: First instalment ₹ 150, last instalment ₹ 40

Total amount to be repaid \(= 1000 + 140 = ₹\ 1140\).

The instalments form an A.P. with \(n = 12\), \(d = -10\), \(S_{12} = 1140\).

\(S_n = \dfrac{n}{2}[2a + (n - 1)d]\)

\(1140 = \dfrac{12}{2}[2a + 11(-10)] = 6(2a - 110)\)

\(190 = 2a - 110 \Rightarrow 2a = 300 \Rightarrow a = 150\)

Last instalment \(t_{12} = a + 11d = 150 - 110 = 40\).

Check: \(\dfrac{12}{2}(150 + 40) = 6 \times 190 = 1140\) ✓.

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

Was this answer helpful?

Asked in: March 2025 board paperA farmer borrows ₹ 1,000 and agrees to repay with a total interest of ₹ 140, in 12 instalments. Each instalment being less than the preceding instalment by ₹ 10. What should be the amount of his first and last instalme…

1 answer

Your answer