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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 \( \alpha \), then \( 2 \alpha \) is [JEE Mains (Feb) 2021]

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Let the terms of the GP be α, αr, αr2, αr3

Therefore from the given condition we have

The sum of the first four terms is

The sum of the reciprocal of the first four term is

Dividing (i) by (ii) we get,

Given that the product of the first three terms is 1

Using result (iii) & (iv) we get

α = 2/3

Putting the value of α in (iv) we get,

r = 3/2

From given

Hence, the value of 2α is 3.

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