JEE Main · 2011 · 10 Apr 2011 · Offline · Sequences and Series

The first three months save $200$ each, a total of $600$. From the fourth month the savings form an arithmetic progression with first term $240$ and common difference $40$. If there are $n$ such later months, their sum is $\frac{n}{2}\left[2(240)+(n-1)40\right]=20n(n+11)$. The grand total is $600+20n(n+11)=11040$, so $20n(n+11)=10440$, i.e. $n^2+11n-522=0$, which factors to give $n=18$. Adding the first three months, the total time is $18+3=21$ months.