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, $a_1>0,a_2,a_3,a_4,a_5\to$G.P.

$3a_2+a_3=2a_4$

Using the formula $a_n=a{r}^{n-1}$ and assuming $a_1=a$ we get, $3ar+a{r}^2=2a{r}^3$

$\Rightarrow 3+r=2r^2$

$\Rightarrow 2r^2-r-3=0$

$\Rightarrow r=-1$ & $r=\frac{3}{2}$

Also, $a_2+a_4=2a_3+1$

$\Rightarrow ar+a{r}^3=2a{r}^2+1$

$\Rightarrow a\left(r+r^3-2r^2\right)=1$

$\Rightarrow a\left(\frac{3}{2}+\frac{27}{8}-\frac{18}{4}\right)=1$

$\Rightarrow a=\frac{8}{3}$

When $r=-1,a=-\frac{1}{4}$  (rejected, $a_1>0$)

So, at $r=\frac{2}{3}, a=\frac{8}{3}$ 

Now $a_2+a_4+2a_5$

$=\frac{8}{3}\times \frac{3}{2}+\frac{8}{3}\times \frac{27}{8}+2\times \frac{8}{3}\times \frac{81}{16}$

$=4+9+27=40$