Question

Let $S_n$ denote the sum of first $n$ terms of an $AP$, If $S_{2n}=3S_n$ then find the ratio $S_{3n}/S_n$.

Answer 1

Let the first term of AP be a and common difference be d

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

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

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

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

$n[2a+2nd-d]=\frac{3n}{2}[2a+nd-d]$

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

$\Rightarrow 2a-nd+d$

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

$\Rightarrow 2a=(n-1)d$ …………$\left(1\right)$

To find: ratio $S_{3n}/S_n$

$\frac{S_{3n}}{S_n}=\frac{\frac{3n}{2}[2a+(3n-1\left)d\right]}{\frac{n}{2}[2a+(n-1\left)d\right]}=\frac{6a+9nd-3d}{2a+\left(nd\right)-d}$

From equation $\left(1\right)$

$\frac{S_{3n}}{S_n}=\frac{3nd-3d+9nd-3d}{nd-d+nd-d}=\frac{6d(2n-1)}{d(2n-1)}=6$

$\Rightarrow \frac{S_{3n}}{S_n}=6$.