Question -
Answer -
Let the three numbersbe a/r, a, ar
So, according to thequestion
a/r + a + ar = 65 …equation (1)
a/r × a × ar = 3375 …equation (2)
From equation (2) weget,
a3 =3375
a = 15.
From equation (1) weget,
(a + ar + ar2)/r= 65
a + ar + ar2 =65r … equation (3)
Substituting a = 15 inequation (3) we get
15 + 15r + 15r2 =65r
15r2 –50r + 15 = 0… equation (4)
Dividing equation (4)by 5 we get
3r2 –10r + 3 = 0
3r2 –9r – r + 3 = 0
3r(r – 3) – 1(r – 3) =0
r = 3 or r = 1/3
Now, the equation willbe
15/3, 15, 15×3 or
15/(1/3), 15, 15×1/3
So the terms are 5,15, 45 or 45, 15, 5
∴ The threenumbers are 5, 15, 45.