Question:
Find the sum of first n natural numbers.
Solution:
The first n natural numbers are:
1, 2, 3, 4...
a = 1, d = 1, Total terms = n
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{n}=\frac{n}{2}[2 \times 1+(n-1) 1]$
$\Rightarrow S_{n}=\frac{n}{2}[2+(n-1) 1]$
$\Rightarrow S_{n}=\frac{n}{2}[n+1]$