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 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.
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