If $S_{n}$ denote the sum of the first $n$ terms of an A.P. If $S_{2 n}=3 S_{n}$, then $S_{3 n}: S_{n}$ is equal to
(a) 4
(b) 6
(c) 8
(d) 10
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).
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