Question

The sums of n terms of three arithmetical progressions 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 $S_1+S_3=2S_2$.

Answer 1

Here $a=1$ for all and $d=1,2,3,$ respectively for $S_1,S_2,S_3,$ and $n=n$ for all.
$S_1+S_3=\left(n/2\right)[2.1+(n-1\left).1\right]+\left(n/2\right)[2.1+(n-1\left).3\right]$
$=\left(n/2\right)[n+1]+\left(n/2\right)[3n-1]$
$=\left(n/2\right)4n=2n^2$
$2S_2=2.\left(n/2\right)[2.1+(n-1\left).2\right]=n.\left(2n\right)=2n^2$
$\therefore S_1+S_3=2.S_2$.