A man saved Rs 16500 in ten years. In each year after the first he saved

Here, we are given that the total saving of a man is Rs 16500 and every year he saved Rs 100 more than the previous year.

So, let us take the first installment as a.

Second installment = 

Third installment = 

So, these installments will form an A.P. with the common difference (d) = 100

The sum of his savings every year 

Number of years (n) = 10

So, to find the first installment, we use the following formula for the sum of n terms of an A.P.,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 10, we get,

$S_{10}=\frac{10}{2}[2(a)+(10-1)(100)]$

$16500=5[2 a+(9)(100)]$

$16500=10 a+4500$

$16500-4500=10 a$

Further solving for a,

$10 a=12000$

$a=R s 1200$

Therefore, man saved ${R s} 1200$ in the first year.