Given, a1>0,a2,a3,a4,a5→G.P.
3a2+a3=2a4
Using the formula an=arn−1 and assuming a1=a we get, 3ar+ar2=2ar3
⇒3+r=2r2
⇒2r2−r−3=0
⇒r=−1 & r=23
Also, a2+a4=2a3+1
⇒ar+ar3=2ar2+1
⇒a(r+r3−2r2)=1
⇒a(23+827−418)=1
⇒a=38
When r=−1,a=−41 (rejected, a1>0)
So, at r=32,a=38
Now a2+a4+2a5
=38×23+38×827+2×38×1681
=4+9+27=40