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 nterms is

विकल्प

•  1/5

•  2/3

• 3/4

• none of these

Answer 1

1/5

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

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

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

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

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

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

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

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

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

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

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

$$=\frac{1}{5}$$