Question

For an A.P.. S${}_{2n}=3S_n$. The value of $\frac{S_{3n}}{S_n}$ is equal to

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is B $6$

For an A.P,$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

Thus, $S_{2n}=n(2a+(2n-1\left)d\right)$

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

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

After cancelling $n$ from both sides, we get

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

Further after solving, we get

$\frac{1}{2}d(n+1)=a$

Therefore, the value of $\frac{S_{3n}}{S_n}$ is

$=3.\frac{2a+(3n-1)d}{2a+(n-1)d}$

$=3.\frac{4nd}{2nd}$

$=6$