Question

The sum of n terms of three arithmetical progression are $S_1$, $S_2$ and $S_3$. The first term of each is unity and the common differences are 1, 2 and 3 respectively. Prove that $S_1$ + $S_3$ = 2$S_2$.

Answer 1

We have,
$S_1$ = Sum of n terms of an AP. with first term 1 and common difference 1

=($\frac{n}{2}$) [2 x 1+(n- 1) 1] =($\frac{n}{2}$) [n+ 1]
$S_2$ = Sum of n terms of an AP. with first term 1 and common difference 2

=($\frac{n}{2}$)[2x1+(n—1)x2]= $n^2$
$S_3$ = Sum of n terms of an AP. with first term 1 and common difference3
=($\frac{n}{2}$) [2 x1+(n-1) $\times$ 3]=($\frac{n}{2}$)(3n-1)

Now, $S_1$ + $S_3$ = ($\frac{n}{2}$) (n + 1) + ($\frac{n}{2}$) (3n — 1)
= 2$n_2$ and $S_2$ = $n^2$

Hence $S_1$ + $S_3$ = 2$S_2$