Question

find four numbers in GP such that their sum is 15 and the sum of their squares is 85.

Answer 1

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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