Question

Let $S_n$ denote the sum of the first n terms of an A.P. If $S_{2n}=3S_n$, then $\frac{S_{3n}}{S_n}=$

• A. 3
• B. 6
• C. 7
• D. 5

Answer 1

Answer: (B)

6

Let a be the first term and d the common difference of the given A.P. Then,

$S_{2n}=3S_n$

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

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

$2a-(3n-3-4n+2)d=0$

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

$2a=(n+1)d$

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

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

$=\frac{12nd}{2nd}=6$