# If the sum of $n,2n\:and\:3n$ terms of an A.P. are $S_1,S_2\:and\:S_3$ respectively, then show that $S_3=3(S_2-S_1)$

Toolbox:
• Sum of $n$ terms of an A.P $=\large\frac{n}{2}$$(2a+(n-1)d) Given that S_1= Sum of n terms =\large\frac{n}{2}$$(2a+(n-1)d)$
$S_2=$Sum of $2n$ terms $=\large\frac{2n}{2}$$(2a+(2n-1)d) and S_3= Sum of 3n terms =\large\frac{3n}{2}$$(2a+(3n-1)d)$ of an A.P.
$\Rightarrow\:S_2-S_1= \big[\large\frac{2n}{2}$$(2a+(2n-1)d)\big]- \big[\large\frac{n}{2}$$(2a+(n-1)d)\big]$
$\qquad\:=\large\frac{n}{2}$$\big[(4a-2a)+(4n-2-n+1)d\big] \qquad\:=\large\frac{n}{2}$$\big[2a+(3n-1)d\big]$
$3(S_2-S_1)=3\large\frac{n}{2}$$\big[2a+(3n-1)d\big]=S_3$
Hence proved.