In the arithmetic progression whose common difference is non-zero,

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 2 n terms to the next 2 n terms is

(a) 1/5

(b) 2/3

(c) 3/4

(d) none of these

Solution:

(a) 1/5

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

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

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

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

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

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

$\Rightarrow 2 a+(n-1) d=0$    ...(1)

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

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

$=\frac{n(n d)}{2 n(3 n d)-n(n d)}$

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

$=\frac{1}{5}$