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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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The sum of first four terms of a geometric progression (G.P.) is 6512 and the sum of their respective reciprocals is 6518. If the product of first three terms of the G.P. is 1, and the third term is α , then 2α is