Question

Four numbers are in GP, if the product of the extremes is 243 & sum of the middle two is 36,find the numbers

Answer 1

The four numbers would be 
a, ar, ar^2, and ar^3 

a(ar^3) = 243 
(a^2)(r^3) = 243 or ar(ar^2) = 243 

ar + ar^2 = 36 
ar(1 + r) = 36 

divide ar(ar^2) = 243 by ar(1+r)=36 
ar^2/(1+r) = 27/4 
4ar^2 = 27 + 27r 
a = (27+27r)/(4r^2) = 27(1+r)/(4r^2) 

plug into: a^2 r^3 = 243 
[ 27(1+r)/(4r^2) ]^2 r^3 = 243 
729(1+r)^2/16r^4 (r^3) = 243 
3(1+r)^2 /(16r) = 1 
3(1+r)^2 = 16r 
3 + 6r + 3r^2 = 16r 
3r^2 - 10r + 3 = 0 
(3r -1)(r - 3) = 0 
r = 3 or r = 1/3 

if r = 3, a = 27(4)/(36) = 3 
the four terms are : 3 ,9,27,81