Question

Let $S_n$ denote the sum of the 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 $6$

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

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

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

Given $S_{2n}$ $=$ 3. $S_n$ $\Rightarrow$ $\frac{2n}{2}[2a-d+2nd]=3$ $\frac{n}{2}[2a-d+nd]\Rightarrow$ $2a-d=nd$ $\to$ (1)
using (1)
$S_{3n}$ $=$ 6$n^{2}d$ and $S_n$ $=$ $n^{2}d$
$S_{3n}$ : $S_n$ $=6$