Question

The sums of first n terms of three A.P.s are $S_1,S_2$ and $S_3$. The first term of each). their common differences are 2, 4 and 6 respectively. Prove that $S_1+S_3=2S_2$ .

Answer 1

As we know,
$S=\frac{n}{2}(2a+(n-1\left)d\right)$
So,

$S_1=\frac{n}{2}(2a+(n-1\left)2\right)$
$S_2=\frac{n}{2}(2a+(n-1\left)4\right)$
$S_3=\frac{n}{2}(2a+(n-1\left)6\right)$

Add $S_{1}and{S}_3$

$S_1+S_3=\frac{n}{2}(2a+(n-1\left)2\right)+\frac{n}{2}(2a+(n-1\left)6\right)$

$=\frac{n}{2}(4a+(n-1\left)8\right)$

$=2\frac{n}{2}(2a+(n-1\left)4\right)$

If you observe closely, the above equation is twice of $S_2$

$\therefore S_1+S_3=2S_2$

Hence proved.