If S_n denote the sum of the first n terms of an A.P. If S_2n = 3S_n , then S_3n : S_n is equal to

$${S_n}$$ = Sum of $$n$$ terms of A.P. and $${S_{2n}}$$ = $$3{S_n}$$
$${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right],$$     $${S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]$$     and $${S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]$$
We know that $${S_{3n}} = 3\left( {{S_{2n}} - {S_n}} \right)$$    and $${S_{2n}} = 3{S_n}$$
$$\eqalign{ & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3\left( {{S_{2n}} - {S_n}} \right)}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3\left( {3{S_n} - {S_n}} \right)}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3 \times 2{S_n}}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{6}{1} \cr & \therefore {S_{3n}}:{S_n} = 6 \cr} $$