Question

Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3n}=3S_{2n},$ then the value of $\frac{S_{4n}}{S_{2n}}$ is

Answer 1

$S_{3n}=\frac{3n}{2}[2a+(3n-1\left)d\right]$
$S_{2n}=n[2a+(2n–1\left)d\right],$
where $a$ is the first term & $d$ is the common difference of A.P.

$S_{3n}=3S_{2n}$
$\Rightarrow \frac{3}{2}[2a+(3n-1\left)d\right]=3[2a+(2n-1\left)d\right]\Rightarrow 2a+(3n-1)d=4a+2(2n–1)d\Rightarrow 2a=(3n-1-4n+2)d\Rightarrow \frac{a}{d}=\frac{1-n}{2}\dots \left(1\right)$

Now, $\frac{S_{4n}}{S_{2n}}=\frac{\frac{4n}{2}[2a+(4n-1\left)d\right]}{\frac{2n}{2}[2a+(2n-1\left)d\right]}$
$=\frac{2[2\left(\frac{1-n}{2}\right)+(4n-1\left)\right]}{[2\left(\frac{1-n}{2}\right)+(2n-1\left)\right]}=\frac{2(1-n+4n-1)}{1-n+2n-1}=\frac{2\left(3n\right)}{n}=6$