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 ______.

Let the four terms of the G.P. be $$a, ar, ar^2, ar^3$$. We are given that their sum is $$\frac{65}{12}$$, the sum of their reciprocals is $$\frac{65}{18}$$, and the product of the first three terms is 1.

The sum of reciprocals is $$\frac{1}{a}\left(1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3}\right) = \frac{1}{a} \cdot \frac{1 + r + r^2 + r^3}{r^3}$$. Dividing the sum by the sum of reciprocals: $$\frac{a(1 + r + r^2 + r^3)}{\frac{1}{a} \cdot \frac{1 + r + r^2 + r^3}{r^3}} = a^2 r^3 = \frac{65/12}{65/18} = \frac{18}{12} = \frac{3}{2}$$.

The product of the first three terms is $$a \cdot ar \cdot ar^2 = a^3 r^3 = 1$$, so $$ar = 1$$, which gives $$a = \frac{1}{r}$$.

Substituting into $$a^2 r^3 = \frac{3}{2}$$: $$\frac{r^3}{r^2} = r = \frac{3}{2}$$. Then $$a = \frac{1}{r} = \frac{2}{3}$$.

The third term is $$\alpha = ar^2 = \frac{2}{3} \cdot \frac{9}{4} = \frac{3}{2}$$. Therefore, $$2\alpha = 2 \cdot \frac{3}{2} = 3$$.

The correct answer is $$3$$.