Question 483278
Let the 4 numbers in GP be
a/r, a, ar, ar^2

Given that sum is 85 and their product is 4096. 

Accordingly....
a/r+a+ar+ar^2=85 & a/r*a*ar*ar^2=4096
a/r(1+r+r^2+r^3)=85 & a^4r^2 = 4096 

Since a^4r^2 = 4096 so a^4= 4096/r^2 = 8^4/r^2
So a = 8/sqrt(r)

Substituting value of a in sum equation leads to ...
(8/sqrt(r)r)(1+r+r^2+r^3)=85
(1+r+r^2+r^3)/(sqrt(r)r)= 85/8
Therefore 
(1+r+r^2+r^3)= 85 and sqrt(r)r = 8
Solving 2nd eqn
r*r^(1/2)= 2^(3/2)
r^(3/2)= 4^(3/2)
so r = 4

Substituting value of r in below eqn and solving for a
a^4r^2 = 4096 
(a^4)(4^2)=4096 
(a^4)*16= 4096 
(a^4)= 4096/16 = 256 = 4^4
So a = 4

Since four numbers in GP are...a/r, a, ar, ar^2
Therefore four numbers in GP are...
4/4, 4, 4*4, 4*4^2
1, 4, 16, 64