Question

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

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

The correct option is B

6

As, $S_{n}2n=3S_n$

$\Rightarrow \frac{2n}{2}[2a+(2n-1\left)d\right]$

$=\frac{3n}{2}[2a+(n-1\left)d\right]$

$\Rightarrow 2[2a+(2n-1\left)d\right]=3[2a+(n-1\left)d\right]$

$\Rightarrow 4a+2(2n-1)d=6a+3(n-1)d$

$\Rightarrow 4a+4nd-2d=6a+3nd-3d$

$\Rightarrow 6a-4a+3nd-3d-4nd+2d=0$

$\Rightarrow 2a-nd-d=0$

$\Rightarrow 2a-(n+1)d=0$

$\Rightarrow 2a+(n+1)d\dots \dots \left(i\right)$

Now,

$\frac{S_{3n}}{S_n}=\frac{\frac{3n}{2}[2a+(3n-1\left)d\right]}{\frac{n}{2}\left[\right(2a+9n-1\left)d\right]}$

$=\frac{3\left[\right(n+1)d+(3n-1\left)d\right]}{\left[\right(n+1)d+(n-1\left)d\right]}$ [using (i)]

$=\frac{3[nd+d+3nd-d]}{[2d+d+nd-d]}$

$=\frac{3\left[4nd\right]}{\left[2nd\right]}$

$=3\times 2=6$

Hence, the correct alternative is option (b).