Let the three numbers be a/r, a, ar
So, according to the question
a/r + a + ar = 38 … equation (1)
a/r × a × ar = 1728 … equation (2)
From equation (2) we get,
a3 = 1728
a = 12.
From equation (1) we get,
(a + ar + ar2)/r = 38
a + ar + ar2 = 38r … equation (3)
Substituting a = 12 in equation (3) we get
12 + 12r + 12r2 = 38r
12r2 – 26r + 12 = 0… equation (4)
Dividing equation (4) by 2 we get
6r2 – 13r + 6 = 0
6r2 – 9r – 4r + 6 = 0
3r(3r – 3) – 2(3r – 3) = 0
r = 3/2 or r = 2/3
Now the equation will be
12/(3/2) = 8 or
12/(2/3) = 18
So the terms are 8, 12, 18
∴ The three numbers are 8, 12, 18