Ramesh saves money each month in an arithmetic progression. He saves Rs $1200$ in the first month and increases the saving by Rs $300$ each subsequent month. After how many months will his total savings be at least Rs $3,00,000$? Also state the total amount saved at the end of that month.

Step-by-step solution

1. Let the first saving be $a=1200$ and common difference be $d=300$. The total after $n$ months is $S_n=\frac{n}{2}[2a+(n-1)d]$.
2. Substitute $a$ and $d$: $S_n=\frac{n}{2}[2400+300(n-1)]=\frac{n}{2}[300n+2100]$. Simplify: $S_n=150n^2+1050n$.
3. Set $S_{n}ge300000$: $150n^2+1050n-300000ge0$. Divide by 150: $n^2+7n-2000ge0$. Solve the quadratic equation $n^2+7n-2000=0$ using the quadratic formula: $n=\frac{-7pm\sqrt{49+8000}}{2}$. $\sqrt{8049}\approx 89.78$, so $n\approx \frac{-7+89.78}{2}\approx 41.39$. Since $n$ must be a whole number of months, the smallest $n$ satisfying the inequality is $42$.
4. Compute $S_{42}$: $S_{42}=150(42{)}^2+1050(42)=150\times 1764+44100=264600+44100=308700$.
5. Thus Ramesh will reach at least Rs 3,00,000 after 42 months, and the total saved then is Rs 3,08,700.

Concept tested: Sum of AP to reach a target amount

Published 4 October 2026 · How this solution was prepared and checked