A man repays a loan of ₹3,250 by paying ₹20 in the first month and increasing the payment by ₹15 every month. The loan is cleared in:

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20 months. The instalments form an arithmetic progression with a = 20 and d = 15. Sₙ = n/2 [40 + 15(n − 1)] = n(15n + 25)/2 = 3250 gives 15n² + 25n − 6500 = 0, i.e. 3n² + 5n − 1300 = 0. Since √(25 + 15600) = 125, n = (−5 + 125)/6 = 20. Check: S₂₀ = 10(40 + 285) = ₹3,250.

A man repays a loan of ₹3,250 by paying ₹20 in the first month and increasing the payment by ₹15 every month. The loan is cleared in:

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