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, ${a}_{1}>0,{a}_{2},{a}_{3},{a}_{4},{a}_{5}\rightarrow$a1​>0,a2​,a3​,a4​,a5​→G.P.

$$3{a}_{2}+{a}_{3}=2{a}_{4}$$3a2​+a3​=2a4​

Using the formula ${a}_{n}=a{r}^{n-1}$an​=arn−1 and assuming ${a}_{1}=a$a1​=a we get, $3ar+a{r}^{2}=2a{r}^{3}$3ar+ar2=2ar3

$$\Rightarrow 3+r=2{r}^{2}$$⇒3+r=2r2

$$\Rightarrow 2{r}^{2}-r-3=0$$⇒2r2−r−3=0

$\Rightarrow r=-1$⇒r=−1 & $r=\frac{3}{2}$r=23​

Also, ${a}_{2}+{a}_{4}=2{a}_{3}+1$a2​+a4​=2a3​+1

$$\Rightarrow ar+a{r}^{3}=2a{r}^{2}+1$$⇒ar+ar3=2ar2+1

$$\Rightarrow a(r+{r}^{3}-2{r}^{2})=1$$⇒a(r+r3−2r2)=1

$$\Rightarrow a(\frac{3}{2}+\frac{27}{8}-\frac{18}{4})=1$$⇒a(23​+827​−418​)=1

$$\Rightarrow a=\frac{8}{3}$$⇒a=38​

When $r=-1,a=-\frac{1}{4}$r=−1,a=−41​ (rejected, ${a}_{1}>0$a1​>0)

So, at $r=\frac{2}{3},a=\frac{8}{3}$r=32​,a=38​

Now ${a}_{2}+{a}_{4}+2{a}_{5}$a2​+a4​+2a5​

$$=\frac{8}{3}\times \frac{3}{2}+\frac{8}{3}\times \frac{27}{8}+2\times \frac{8}{3}\times \frac{81}{16}$$=38​×23​+38​×827​+2×38​×1681​

$$=4+9+27=40$$=4+9+27=40