Question

Derive the formula for the sum of the first $n$ natural numbers in harmonic progression.

Answer 1

The sequence of the first $n$ natural numbers is $1,2,3,4,5,...,n-1,n$.
Here $a_1=1,d=1$
We know that $S_n=a_1+a_2+a_3+....a_n=(a_1+\frac{(n-1)d}{2})\times n$
$S_n=(1+\frac{(n-1)1}{2})\times n$
$S_n=\frac{n(n+1)}{2}$
The sum of first $n$ natural numbers in H.P. $=$ $\frac{2}{n(n+1)}$.