A number theory problem by Ram Prasanth

Sum of natural numbers from 1 upto $n$ is

$\frac{n(n+1)}{2}=\frac{100 \times 101}{2}=50 \times 101 = \boxed{5050}$

(100+1) + (99+2) ..... + (50+49) = 101 * a
a=100/2=50 since two numbers in each groups
101*50=5050

Arithmetic Progrressive : $a = 1 , n = 100, d = 1$

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

$S_{100} = \frac{1}{2}100 (2 . 1 + (100 - 1)1)$

$S_{100} = 50 . 101 = 5050$