Question

if sum of three numbers in G.P is 38 and their product is 1728, find the numbers

Answer 1

If we use r to denote the common ratio, 
and a to denote the first number, 
then they are 
a, ar, ar^2 
so their sum is a + ar + ar^2 
and their product is a^3 r^3. 

Plainly, the cube root of the product is ar, the second number: 12 

So the sum of the first and last numbers is 
a + ar^2 = 38 - 12 = 26 
a + 12r = 26 [substituting 12 for ar in the second term] 
a = 26 - 12r 

Now we can perform a substitution: 
a (1 + r^2) = 26 
(26 - 12r) (1 + r^2) = 26 
26 - 12r + 26r^2 - 12r^3 = 26 
12r^3 - 26r^2 + 12r = 0 

Since r cannot be 0, we can simplify this by dividing each side by 2r: 
6r^2 - 13r + 6 = 0 
(2r - 3) (3r - 2) = 0 
which has two solutions: 
r = 3/2 and r = 2/3 

If r = 3/2, the three numbers are 8, 12, 18, in that order; 
If r = 2/3, the three numbers are 18, 12, 8, in that order. 
Either case satisfies the description, and the three numbers are the same in both.