A man saved ₹66000 in 20 years.

Solution:

As, in each succeeding year after the first year he saved ₹200 more than what he saved in the previous year.
So, the savings of each year are in A.P.

We have,

the total savings of the man in 20 years, S20 = ₹66000 and

the difference of his savings in each succeeding year, d = ₹200

Let his savings in the first year be a.

Now,

$S_{20}=66000$

$\Rightarrow \frac{20}{2}[2 a+(20-1) d]=66000$

$\Rightarrow 10[2 a+19 \times 200]=66000$

$\Rightarrow 2 a+3800=\frac{66000}{10}$

$\Rightarrow 2 a=6600-3800$

$\Rightarrow a=\frac{2800}{2}$

$\therefore a=1400$

So, he saved ₹1400 in the first year.