# If Sr denotes the sum of the first r terms of an A.P.

Question:

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

Solution:

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.

= 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).