Question

If $S_1,S_2,S_3$ are the sums of the terms of n three series in A.P., the first term of each being 1 and the respectively common difference being 1, 2, 3; prove the
$S_1+S_3=2S_2$

Answer 1

$a_1=1,a_2=1,a_3=1$

$d_1=1,d_2=2,a_d=3$

$s_1=\frac{n}{2}[2+(n-1\left)\right]=\frac{n}{2}[n+1]$

$s_2=\frac{n}{2}[2+(n-1\left)2\right]=\frac{n}{2}\times 2n=n^2$

$s_3=\frac{n}{2}[2+(n-1\left)3\right]=\frac{n}{2}[3n+1]$

$S_1+S_3=2S_2$

$\frac{n}{2}[n+1]+\frac{n}{2}[3n+1]={2n}^2$

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

$\frac{{4n}^2}{2}={2n}^2$ $\left(proved\right)$