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, $S_{2n}=3S_n$
$\frac{2n}{2}[2a+(2n-1)d]=3\cdot \frac{n}{2}[2a+(n-1)d]$
Where, $a$ and $d$ are first term and common difference of an AP respectively.
$\Rightarrow 4a+2(2n-1)d=6a+3(n-1)d$
$\Rightarrow 2a+(3n-3-4n+2)d=0$
$\Rightarrow 2a+(-n-1)d=0$
$\Rightarrow 2a+(n+1)(-d)=0$
$\Rightarrow 2a=(n+1)d\dots (i)$
Now, $\frac{S_{3n}}{S_n}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}$
$=\frac{3[(n+1)d+(3n-1)d]}{[(n+1)d+(n1)d]}$ [From Eq. (i)]
$=\frac{3[(n+1+3n-1)d]}{(n+1+n-1)d}=\frac{3(4nd)}{(2nd)}=6$
$\Rightarrow S_{3n}=6S_n$
On compare with, $S_{3n}=k{S}_n$
$\Rightarrow k=6$