SOLUTION: Find four numbers in GP whose sum is 85 and their product is 4096

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Question 483278: Find four numbers in GP whose sum is 85 and their product is 4096
Answer by mathstutor494(120)   (Show Source): You can put this solution on YOUR website!
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







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