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Q.

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then 2 $α$ is

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a

2

b

1

c

3

d

4

answer is C.

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Detailed Solution

$\frac{a + ar + {ar}^{2} + {ar}^{3}}{\frac{1}{a} + \frac{1}{ar} + \frac{1}{{ar}^{2}} + \frac{1}{{ar}^{3}}} = \frac{\frac{65}{12}}{\frac{65}{18}} a^{2} r^{3} = \frac{18}{12} = \frac{3}{2} a \cdot ar \cdot {ar}^{2} = 1 \Rightarrow a^{3} r^{3} = 1 \Rightarrow ar = 1 \Rightarrow a = \frac{1}{r} \frac{1}{r^{2}} \cdot r^{3} = \frac{3}{2} \Rightarrow r = \frac{3}{2} a = \frac{2}{3} α = a_{3} = {ar}^{2} \Rightarrow α = \frac{2}{3} \times \frac{9}{4} = \frac{3}{2}$

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