Solution:

As per the question:
A man saved ₹$16500$ in ten years

Let be his savings in the first year be ₹ $x$

Every year his savings increased by ₹ 100.

Therefore,

A.P will be $x$, $100 + x$, $200 + x \dots \dots$.

In which, $x$ is first term and

Common difference, $d = 100 + x – x = 100$

It is known that, $S_n$ is the sum of $n$ terms of an A.P

Using the formula,

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

Where, the first term is $a$, the common difference is $d$ and $n$ is number of terms in an A.P.
Given that,
$\begin{array}{l}
S_{n}=16500 \text { and } n=10 \\
S_{10}=10 / 2[2 x+(10-1) 100] \\
16500=5\{2 x+9(100)\} \\
16500=5(2 x+900) \\
16500=10 x+4500 \\
-10 x=4500-16500 \\
-10 x=-12000 \\
x=12000 / 10 \\
=1200
\end{array}$
As a result, his saving in the first year is ₹ 1200 .