Question

If $S_{2n}=3S_n$ and $S_{5n}=k{S}_{3n}$, where $S_n$ is the sum of $n$ terms of an A.P., then the value of $k$ is

• A. $2$
• B. $2.5$
• C. $3$
• D. $3.5$

Answer 1

The correct option is B $2.5$

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

$\frac{S_{5n}}{S_{3n}}=\frac{\frac{5n}{2}[2a+(5n-1\left)d\right]}{\frac{3n}{2}[2a+(3n-1\left)d\right]}\Rightarrow k=\frac{5\left[\right(n+1)d+(5n-1\left)d\right]}{3\left[\right(n+1)d+(3n-1\left)d\right]}\Rightarrow k=\frac{5\times 6nd}{3\times 4nd}\Rightarrow k=2.5$