Question

In the arithmetic progression whose common difference is non - zero, the sum of first 3 n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2n terms to the next 2n terms is

• A. $\frac{1}{5}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{5}$

$S_{3n}=S_{4n}-S_{3n}$

$\Rightarrow 2S_{3n}=S_{4n}$

$\Rightarrow 2\times \frac{3n}{2}\{2a+(3n-1\left)d\right\}=\frac{4n}{2}\{2a+(4n-1\left)d\right\}$

$\Rightarrow 3\{2a+(3n-1\left)d\right\}=2\{2a+(4n-1\left)d\right\}$

$\Rightarrow 6a+9nd-3d=4a+8nd-2d$

$\Rightarrow 2a+nd-d=0$

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

Required ratio : $\frac{S_{2n}}{S_{4n-S_{2n}}}$

$\frac{S_{2n}}{S_{4n}-S_{2n}}$

$=\frac{\frac{2n}{2}\{2a+(2n-1\left)d\right\}}{\frac{4n}{2}\{2a+(4n-1\left)d\right\}-\frac{2n}{2}\{2a+(2n-1\left)d\right\}}$

$=\frac{n\left(nd\right)}{2n\left(3nd\right)-n\left(nd\right)}$

$=\frac{1}{6-1}$

$=\frac{1}{5}$