# The sum of three numbers in G.P. is 21, and the sum of their squares is 189. Find the numbers.

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

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