**Question:**

A man saves ₹4000 during the first year, ₹5000 during the second year and in this way he increases his savings by ₹1000 every year. Find in what time his savings will be ₹85000.

**Solution:**

A Man saves some amount of money every year.

In the first year, he saves Rs.4000.

In the next year, he saves Rs. 5000 .

Like this, he increases his savings by Rs. 1000 ever year.

Given a total amount of Rs. 85000 is saved in some 'n' years.

According to the above information the savings in every year are in Arithmetic Progression.

First year savings $=a=$ Rs 4000

Increase in every year savings $=d=R s .1000$

Total savings $\left(\mathrm{s}_{\mathrm{n}}\right)=\mathrm{Rs} .85000$

Sum of $n$ terms in A.P $=\frac{n}{2}[2 \times a+(n-1) \times d]$

$s_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 4000+(\mathrm{n}-1) \times 1000]$

$85000=\frac{n}{2}[8000+(n-1) \times 1000]$

$n^{2}+7 \times n-170=0$

$(n+17) \times(n-10)=0$

$n=-17$ or $n=10$

Since the number of years cannot be negative, $n=10$.

After 10 years his savings will become Rs. 85000 .

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