Find the sum of first n natural numbers.

The first n natural numbers are 1, 2, 3, 4, 5, ..., n.

Here, a = 1 and d = (2 - 1) = 1

Sum of $n$ terms of an AP is given by

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

$=\left(\frac{n}{2}\right) \times[2 \times 1+(n-1) \times 1]$

$=\left(\frac{n}{2}\right) \times[2+n-1]=\left(\frac{n}{2}\right) \times(n+1)=\frac{n(n+1)}{2}$