Question

The sums of first n terms of three arithmetic progressions are $S_1,S_2$ and $S_3$ respectively. The first term of each A.P. is 1 and their common difference are 1, 2 and 3 respectively. Prove that $S_1+S_3=2S_2$.

Answer 1

Given, first term of each A.P. $\left(a\right)=1$.

and their common difference are 1,2 and 3.

$\therefore S_1=\frac{n}{2}[2a+(n-1\left)d_1\right]$

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

$S_2=\frac{n}{2}[2a+(n-1\left)d_2\right]$

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

and $S_3=\frac{n}{2}[2a+(n-1\left)d_3\right]$

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

Now, $S_1+S_3=\frac{n}{2}(n+1)+\frac{n}{2}(3n-1)$

= $\frac{n}{2}(n+1+3n-1)=4n\times \frac{n}{2}=2n^2$

= $2S_2$

$\therefore S_1+S_3=2S_2$

Hence Proved.