Question

Let $S_{n}$ denote the sum of first $n$ terms of an $A.P$. and $S_{2n} = 3S_{n}$. If $S_{3n} =k S_{n}$, then the value of $k$ is equal to

• A. $4$
• B. $5$
• C. $6$
• D. $7$
• E. $8$

Answer 1

Answer: (C)

Given, $S_{2 n}=3 S_{n}$
$\frac{2 n}{2}[2 a+(2 n-1) d]=3 \cdot \frac{n}{2}[2 a+(n-1) d]$
Where, $a$ and $d$ are first term and common difference of an AP respectively.
$\Rightarrow \, 4 a+2(2 n-1) d=6 a+3(n-1) d $
$ \Rightarrow \, 2 a+(3 n-3-4 n+2) d=0 $
$ \Rightarrow \, 2 a+(-n-1) d=0 $
$ \Rightarrow \, 2 a+(n+1)(-d)=0 $
$ \Rightarrow \, 2 a=(n+1) d \,\,\,\,\,\,\dots(i)$
Now, $ \frac{S_{3 n}}{S_{n}}=\frac{\frac{3 n}{2}[2 a+(3 n-1) d]}{\frac{n}{2}[2 a+(n-1) d]} $
$=\frac{3[(n+1) d+(3 n-1) d]}{[(n+1) d+(n \quad 1) d]}$ [From Eq. (i)]
$=\frac{3[(n+1+3 n-1) d]}{(n+1+n-1) d}=\frac{3(4 n d)}{(2 n d)}=6 $
$\Rightarrow \, S_{3 n}=6 S_{n}$
On compare with, $S_{3 n}=k S_{n}$
$\Rightarrow \, k=6$