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

Suppose his total savings is $₹11040$ after $n$ months. His savings are
                      $200, 200, 200, 240, 280, ....$upto $n$ terms
Note that
                     $200, 240, 280, ....$ upto $(n-2)$ terms is an A.P. whose sum is $10640.$
Thus,     $\frac{1}{2}(n-2)\left[2\right(200)+(n-3\left)\right(40\left)\right]=10640$
        $\begin{array}{l}\Rightarrow (n-2)(n+7)=532\\ \Rightarrow n^2+5n-546=0\Rightarrow n=-26,21\end{array}$
As $n>0, n=21.$