Answer: (3)

Solution

$a, ar, ar^2, ar^3$

$a + ar + ar^2 + ar^3 = \frac{65}{12}$ …(1)

$\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{65}{18}$ …(2)

$\frac{(1)}{(2)} \Rightarrow a^2r^3 = \frac{18}{12} = \frac{3}{2}$

$a^3r^3 = 1 \Rightarrow a\left(\frac{3}{2}\right)^{1/3}… $

Actually: $a^3r^3 = 1 \Rightarrow ar = 1 \Rightarrow a = \frac{1}{r}$

$\Rightarrow r^3 = \frac{3}{2}… $ Wait. Let me redo.

$\frac{a(1+r+r^2+r^3)}{(1/a)(1+1/r+1/r^2+1/r^3)} = \frac{65/12}{65/18}$

$a^2r^3 = \frac{3}{2}$

Product of first 3 terms: $a \cdot ar \cdot ar^2 = a^3r^3 = 1$

So $ar = 1$, $a = 1/r$

$\frac{1}{r^2} \cdot r^3 = \frac{3}{2} \Rightarrow r = \frac{3}{2}$

$\alpha = ar^2 = \frac{1}{r} \cdot r^2 = r = \frac{3}{2}$

$2\alpha = 3$