If Sn denote the sum of the first n terms of an A.P.

Here, we are given an A.P. whose sum of $n$ terms is $S_{n}$ and $S_{2 n}=3 S_{n}$.

We need to find $\frac{S_{3 n}}{S_{n}}$.

Here we use the following formula for the sum of n terms of an A.P.,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, first we find S3n,

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

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

Similarly,

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

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

Also,

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

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

Now, $S_{2 n}=3 S_{n}$

So, using $(2)$ and $(3)$, we get,

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

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

On further solving, we get,

$2(2 a+2 n d-d)=3(2 a+n d-d)$

$4 a+4 n d-2 d=6 a+3 n d-3 d$

$2 a=n d+d$ ......(4)

So,

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

Taking $\frac{n}{2}$ common, we get,

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

$=\frac{3(n d+d+3 n d-d)}{(n d+d+n d-d)}$(Using 4)

$=\frac{3(4 n d)}{2 n d}$

$=6$

Therefore, $\frac{S_{3 n}}{S_{n}}=6$

Hence, the correct option is (b).