A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent 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.
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a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms
\( = \frac{n}{2}\) [400 + (n – 1)40] = 11040 – 400
200n + 20n2 – 20n = 10640
20n2 + 180 n – 10640 = 0
n2 + 9n – 532 = 0
(n + 28) (n – 19) = 0
n = 19
\ no. of months = 19 + 2 = 21 .