If $S_{r}$ denotes the sum of the first $r$ terms of an A.P. Then, $S_{3 n}$ : $\left(S_{2 n}-S_{n}\right)$ is
(a) n
(b) 3n
(c) 3
(d) none of these
Here, we are given an A.P. whose sum of $r$ terms is $S_{r}$. We need to find $\frac{S_{3 n}}{S_{2 m}-S_{n}}$.
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)
So, using $(1),(2)$ and $(3)$, we get,
$\frac{S_{3 n}}{S_{2 n}-S_{n}}=\frac{\frac{3 n}{2}[2 a+3 n d-d]}{\frac{2 n}{2}[2 a+2 n d-d]-\frac{n}{2}[2 a+n d-d]}$
Taking $\frac{n}{2}$ common, we get,
$\frac{S_{3 n}}{S_{2 n}-S_{n}}=\frac{3[2 a+3 n d-d]}{2[2 a+2 n d-d]-[2 a+n d-d]}$
$=\frac{3[2 a+3 n d-d]}{4 a+4 n d-2 d-2 a-n d+d}$
$=\frac{3[2 a+3 n d-d]}{2 a+3 n d-d}$
$=3$
Therefore, $\frac{S_{3 n}}{S_{2 n}-S_{n}}=3$
Hence, the correct option is (c).