A man saved ₹33000 in 10 months. In each month after the first, he saved ₹100 more than he did in the preceding month

Let the money saved by the man in the first month be ₹a.

It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹100.

∴ d = ₹100

Number of months, n = 10

Sum of money saved in 10 months, S10 = ₹33,000

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get

$S_{10}=\frac{10}{2}[2 a+(10-1) \times 100]=33000$

$\Rightarrow 5(2 a+900)=33000$

$\Rightarrow 2 a+900=6600$

$\Rightarrow 2 a=6600-900=5700$

$\Rightarrow a=2850$

Hence, the money saved by the man in the first month is ₹2,850.