Let the three numbers be a, ar, and ar²

∴ According to the question

⇒ a + ar + ar² = 21

a(1 + r + r²) = 21

Squaring both sides we get,

a²(1 + r + r²)² = (21)²….(1)

And from the second condition,

a² + a²r² + a²r⁴ = 189

a²(1 + r² + r⁴) = 189……(2)

Dividing both the equations we get,

Cross multiplying we get,

3 + 3r + 3r² = 7r² – 7r + 7

4r² – 10r + 4 = 0

2r² – 5r + 2 = 0

Factorizing the quadratic equation such that, on multiplication, we get 4 and on the addition, we get 5. So,

2r² – (4r + r) + 2 = 0

2r(r – 2) –1(r – 2) = 0

(2r – 1)(r – 2) = 0

r = 1/2 , r = 2

Putting the value of r in equation 2 we get,

At r = 2,

a²(1 + r² + r⁴) = 189

a²(1 + 4 + 16) = 189

a²

a² = 9

a = ±3

At r = 1/2

a² = 9 × 16

a = 3 × 4 = 12

The numbers are:

1, 9, 81 or 81, 9, 1