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 :

Let the time taken to save Rs 11040 be (n + 3) months.

for first three months, he saves Rs 200 each month.

In (n + 3) months,

3 \(\times\) 200 + \(n\over 2\) { 2(240) + (n – 1) \(\times\) 40 } = 11040

\(\implies\)  600 + \(n\over 2\) {40(12+ n – 1)} = 11040

\(\implies\)  600 + 20n(n + 11) = 11040

\(\implies\) 30 + \(n^2\) + 11n = 552

\(\implies\)  \(n^2\) + 11n – 552 = 0

\(\implies\)  \(n^2\) + 29n – 18n – 552 = 0

\(\implies\)  n(n + 29)  – 18(n +29) = 0

\(\implies\)   (n – 18)(n + 29) = 0

\(\therefore\)   n = 18,  n = -29 (neglect)

\(\therefore\)   Total time = (n + 3) = 21 months


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