Question

Let $S_n$ denote the sum of first n terms of an AP and $3S_n=S_{2n}$. What is $S_{3n}:S_n$ equal to?

• A. $4:1$
• B. $6:1$
• C. $8:1$
• D. $10:1$

Answer 1

The correct option is B $6:1$

We know:

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

$\therefore$ According to problem

$\Rightarrow 3S_n=S_{2n}$

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

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

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

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

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

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

$=\frac{3d.4n}{d.2n}$

$=\frac{6}{1}$