Question

Let the sum of $n,2n,3n$ terms of an A.P. be $S_1,S_2$ and $S_3$, respectively, show that $S_3=3(S_2-S_1)$.

Answer 1

Let $a$ and $d$ be the first term and the common difference of the A.P. respectively.

Therefore,

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

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

$S_3=\frac{3n}{2}[2a+(3n-1\left)d\right]...\left(3\right)$

From (1) and (2), we obtain

$S_2-S_1=n[2a+(2n-1\left)d\right]-\frac{n}{2}[2a+(n-1\left)d\right]$

$=n\left\{\frac{4a+4nd-2d-2a-nd+d}{2}\right\}$

$=n\left[\frac{2a+3nd-d}{2}\right]$

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

$\therefore 3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1\left)d\right]=S_3$$\left[From\right(3\left)\right]$ $\left[henceproved\right]$