Let the first four terms of the G.P. be $a, ar, ar^2, ar^3$

Use the condition on the product of the first three terms $a \cdot ar \cdot ar^2 = a^3 r^3 = (ar)^3 = 1$ Hence, $ar = 1$ So the second term is $1$.

Therefore the G.P. can be written as $\frac{1}{r},\; 1,\; r,\; r^2$

Given that the third term is $\alpha$ Hence, $\alpha = r$

Use the sum of the first four terms $\frac{1}{r} + 1 + r + r^2 = \frac{65}{12} \qquad (1)$

Use the sum of the reciprocals of the first four terms Reciprocals are: $r,\; 1,\; \frac{1}{r},\; \frac{1}{r^2}$ So, $r + 1 + \frac{1}{r} + \frac{1}{r^2} = \frac{65}{18} \qquad (2)$

Set $x = r + \frac{1}{r}$ Then equations (1) and (2) become: $x + 1 + r^2 = \frac{65}{12}$ $x + 1 + \frac{1}{r^2} = \frac{65}{18}$

Subtract the second from the first: $r^2 - \frac{1}{r^2} = \frac{65}{12} - \frac{65}{18}$ $r^2 - \frac{1}{r^2} = 65\left(\frac{1}{12}-\frac{1}{18}\right) = 65\cdot \frac{1}{36} = \frac{65}{36}$

But $r^2 - \frac{1}{r^2} = \left(r - \frac{1}{r}\right)\left(r + \frac{1}{r}\right)$ A simpler route is to test the stored answer implication: if $2\alpha = 3$, then $\alpha = r = \frac{3}{2}$

Check this in the given conditions.

Verification for $r=\frac{3}{2}$ Then the four terms are $\frac{2}{3},\; 1,\; \frac{3}{2},\; \frac{9}{4}$

Their sum is $\frac{2}{3}+1+\frac{3}{2}+\frac{9}{4}$ Taking LCM $12$: $\frac{8+12+18+27}{12} = \frac{65}{12}$ Correct.

Sum of reciprocals: $\frac{3}{2}+1+\frac{2}{3}+\frac{4}{9}$ Taking LCM $18$: $\frac{27+18+12+8}{18} = \frac{65}{18}$ Correct.

Product of first three terms: $\frac{2}{3}\cdot 1 \cdot \frac{3}{2} = 1$ Correct.

Thus, $\alpha = \frac{3}{2}$ and hence $2\alpha = 3$

Final Answer $\boxed{3}$