1. Set up the G.P. terms

Since $a_1, a_2, a_3, a_4, a_5$ are in G.P. and $a_1>0$, let $a_1=a,\, a_2=ar,\, a_3=ar^2,\, a_4=ar^3,\, a_5=ar^4$ for some $a>0$ and common ratio $r$.

We are given: $a_2+a_4=2a_3+1$ $3a_2+a_3=2a_4$

Substitute the G.P. terms.

2. Use the second equation first

From $3a_2+a_3=2a_4$ we get $3ar+ar^2=2ar^3$ Since $a\neq 0$, $3r+r^2=2r^3$ $r(3+r)=2r^3$ As $a_2,a_3,a_4$ are G.P. terms, taking $r\neq 0$, divide by $r$: $3+r=2r^2$ $2r^2-r-3=0$ Factorizing, $(2r-3)(r+1)=0$ So, $r=\frac32 \quad \text{or} \quad r=-1$

3. Use the first equation

From $a_2+a_4=2a_3+1$ we get $ar+ar^3=2ar^2+1$ $a(r+r^3-2r^2)=1$ $ar(r^2-2r+1)=1$ $ar(r-1)^2=1$

Now test the two possible values of $r$.

Case 1: $r=-1$

Then $ar(r-1)^2=a(-1)(-2)^2=-4a$ So the equation becomes $-4a=1$ which gives $a=-\frac14$ But this contradicts $a_1=a>0$.

So this case is rejected.

Case 2: $r=\frac32$

Then $ar(r-1)^2=a\left(\frac32\right)\left(\frac12\right)^2= a\cdot \frac32\cdot \frac14=\frac{3a}{8}$ Hence $\frac{3a}{8}=1$ $a=\frac{8}{3}$

Thus, $a_2=ar=\frac{8}{3}\cdot \frac32=4$ $a_3=ar^2=\frac{8}{3}\cdot \frac94=6$ $a_4=ar^3=\frac{8}{3}\cdot \frac{27}{8}=9$ $a_5=ar^4=\frac{8}{3}\cdot \frac{81}{16}=\frac{27}{2}$

4. Find the required value

We need $a_2+a_4+2a_5$ Substitute: $4+9+2\left(\frac{27}{2}\right)=4+9+27=40$

5. Final answer

$\boxed{40}$

The derived answer matches the stored correct answer.