A man saved $\text{Rs. }16,500$ in $10$ years. In each year after the first year he saved $\text{Rs. }100$ more than he did in the preceding year. How much did he save in the first year?

The savings over the $10$ years form an Arithmetic Progression (AP) where:
* Total number of years ($n$) $=10$
* Common difference ($d$) $=100$
* Sum of $10$ terms ($S_{10}$) $=16500$

The formula for the sum of an AP is:

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

Substitute the given values:

$$16500=\frac{10}{2}[2a+(10-1)100]$$

$$16500=5[2a+900]$$

Divide both sides by $5$:

$$3300=2a+900$$

$$2a=3300-900=2400$$

$$a=\frac{2400}{2}=1200$$

Thus, he saved $\text{Rs. }1200$ in the first year.