If $$a_1(> 0), a_2, a_3, a_4, a_5$$ are in a G.P., $$a_2 + a_4 = 2a_3 + 1$$ and $$3a_2 + a_3 = 2a_4$$, then $$a_2 + a_4 + 2a_5$$ is equal to ______

Given $$a_1 (> 0), a_2, a_3, a_4, a_5$$ are in G.P. with $$a_2 + a_4 = 2a_3 + 1$$ and $$3a_2 + a_3 = 2a_4$$. Find $$a_2 + a_4 + 2a_5$$.

First, let $$a_1 = a$$ and the common ratio be $$r$$, so that $$a_2 = ar,\quad a_3 = ar^2,\quad a_4 = ar^3,\quad a_5 = ar^4$$.

Next, using the condition $$a_2 + a_4 = 2a_3 + 1$$ yields $$ar + ar^3 = 2ar^2 + 1$$. Therefore $$ar(1 + r^2 - 2r) = 1 \implies ar(r-1)^2 = 1 \quad \cdots (1)$$.

Similarly, the equation $$3a_2 + a_3 = 2a_4$$ gives $$3ar + ar^2 = 2ar^3$$. Since $$ar \neq 0$$, it follows that $$3 + r = 2r^2 \implies 2r^2 - r - 3 = 0$$. Factorising, $$(2r - 3)(r + 1) = 0\implies r = \frac{3}{2}$$ or $$r = -1$$.

Now, if $$r = \frac{3}{2}$$, substitution into (1) gives $$a \cdot \frac{3}{2} \cdot \left(\frac{1}{2}\right)^2 = 1 \implies a \cdot \frac{3}{2} \cdot \frac{1}{4} = 1 \implies a = \frac{8}{3}$$. Checking $$a_1 > 0$$: $$a = \frac{8}{3} > 0$$ $$\checkmark$$.

On the other hand, if $$r = -1$$, then (1) becomes $$a(-1)((-1)-1)^2 = a(-1)(4) = -4a = 1 \implies a = -\frac{1}{4} < 0$$ $$\times$$, which contradicts $$a_1 > 0$$.

Substituting $$a = \frac{8}{3}$$ and $$r = \frac{3}{2}$$, we find $$a_2 = \frac{8}{3}\cdot\frac{3}{2} = 4$$, $$a_4 = \frac{8}{3}\cdot\frac{27}{8} = 9$$, and $$a_5 = \frac{8}{3}\cdot\frac{81}{16} = \frac{27}{2}$$. Therefore, $$a_2 + a_4 + 2a_5 = 4 + 9 + 27 = 40$$.

Hence the answer is $$40$$.