101+102+103..........200 ka maan kya hoga

This is an arithmetic progression with the first term $a_1 = 101$, the last term $a_n = 200$, and a common difference $d = 1$.

The formula for the $n$-th term of an arithmetic progression is $a_n = a_1 + (n-1)d$.
Substitute the known values: $200 = 101 + (n-1) \times 1$.
$200 - 101 = n - 1$.
$99 = n - 1$.
$n = 99 + 1 = 100$.
There are 100 terms in the series.

The formula for the sum of an arithmetic progression is $S_n = \frac{n}{2}(a_1 + a_n)$.
Substitute the values: $S_{100} = \frac{100}{2}(101 + 200)$.
$S_{100} = 50(301)$.
$S_{100} = 15050$.

The sum of the series is 15050.