Ramesh saves money each month in an arithmetic progression. He saves Rs $2000$ in the first month and each subsequent month he saves Rs $500$ more than the previous month. After how many months will his total savings be at least Rs $1,20,000$? Also state the total amount saved at the end of that month.

Step-by-step solution

1. Let the first term $a=2000$ and common difference $d=500$.
2. The sum after $n$ months is $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.
3. Substitute $a$ and $d$: $S_n=\frac{n}{2}\left[4000+500(n-1)\right]=\frac{n}{2}(500n+3500)$.
4. Simplify: $S_n=250n^2+1750n$.
5. Set $S_n\ge 120000$: $250n^2+1750n-120000\ge 0$.
6. Divide by $250$: $n^2+7n-480\ge 0$.
7. Solve the quadratic equation $n^2+7n-480=0$:
8. Discriminant $D=7^2+4\times 480=49+1920=1969$, $\sqrt{D}\approx 44.38$.
9. Positive root $n=\frac{-7+44.38}{2}\approx 18.69$.
10. Since $n$ must be an integer and the sum must reach at least the target, take the next whole number: $n=19$ months.
11. Compute the total saved in $19$ months: $S_{19}=250\times 19^2+1750\times 19$.
12. $19^2=361$, so $250\times 361=90250$ and $1750\times 19=33250$.
13. Add: $90250+33250=1,23,500$.
14. Hence, after $19$ months Ramesh will have saved Rs $1,23,500$, which exceeds Rs $1,20,000$.

Concept tested: sum of n terms & inequality

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