Engineering
Mathematics
Arithmetic Progression and Its Properties
Question

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.

19 months

21 months

20 months

18 months

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Solution

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 .