Question -
Answer -
Let the three numbersbe a/r, a, ar
So, according to thequestion
a/r + a + ar = 38 …equation (1)
a/r × a × ar = 1728 …equation (2)
From equation (2) weget,
a3 =1728
a = 12.
From equation (1) weget,
(a + ar + ar2)/r= 38
a + ar + ar2 =38r … equation (3)
Substituting a = 12 inequation (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 willbe
12/(3/2) = 8 or
12/(2/3) = 18
So the terms are 8,12, 18
∴ The threenumbers are 8, 12, 18