A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the sa - Mathematics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after:

Solution & Step-by-step Explanation

Savings for first 3 months: 200, 200, 200 . Sum = 600 .From 4th month onwards, it's an AP: a = 240, d = 40 .Let n be the number of months after the first 3 months.Total saving = 600 + \frac{n}{2} [2 (240) + (n - 1) 40] = 11040 . \frac{n}{2} [480 + 40 n - 40] = 10440 n /2 [440 + 40 n] = 10440 ⟹ n (220 + 20 n) = 10440 20 n^{2} + 220 n - 10440 = 0 ⟹ n^{2} + 11 n - 522 = 0 .Solving for n : n = \frac{- 11 \pm 121 + 2088}{2} = \frac{- 11 \pm 47}{2} = 18 .Total months = 3 + 18 = 21 .

Try it yourself before checking the explanation above.

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after:

19 months

20 months

21 months

18 months

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