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.