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

(b) : Let it happens after n months 3 x 200+(n-3 / 2)\2 x 240+(n-4) 40\ =11040 \ => ((n-3 / 2))(480+40 n-160)=11040-600=10440 \ => n^2+5 n-546=0 \ => 0 Therefore n=21

A man saves ₹ 200 in each of the first three months of ... | Sequence & Series (Mathematics)

(b) : Let it happens after n months $\\3 \times 200+\frac{n-3}{2}\{2 \times 240+(n-4) 40\} =11040$ $\\ \Rightarrow\left(\frac{n-3}{2}\right)(480+40 n-160)$ =11040-600=10440 $\\ \Rightarrow n^2+5 n-546=0 \\ \Rightarrow 0 \therefore n=21$

Explanation

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