Question

Sum of $n$ terms of three A.P. are $S_1,S_2,S_3$ respectively. If first term of each progression is $1$ and common differences are $1,2,3$ respectively, then prove that:
$S_1+S_3=2S_2$

Answer 1

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

Using the above formula, we get:

$S_1=\frac{n+n^2}{2}$

Similarly,

$S_2=n^2$

and

$S_3=\frac{3n^2-n}{2}$

$S_1+S_3$

$=\frac{n+n^2}{2}+\frac{3n^2-n}{2}$

$=\frac{n+n^2+3n^2-n}{2}$

$=\frac{4n^2}{2}$

$=2n^2$

$=2S_2$

$\therefore S_1+S_3=2S_2$