Let the sum of $n, 2 n, 3 n$ terms of an $A.P.$ be $S_{1}, S_{2}$ and $S_{3},$ respectively, show that $S_{3}=3\left(S_{2}-S_{1}\right)$

Let $a$ and $b$ be the first term and the common difference of the $A.P.$ respectively. Therefore,

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

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

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

From $(1)$ and $(2),$ we obtain

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

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

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

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

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

Hence, the given result is proved.