Q. 93.5( 13 Votes )

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

Answer :

Let the three numbers be a, ar, and ar^{2}

∴ According to the question

⇒ a + ar + ar^{2} = 21

a(1 + r + r^{2}) = 21

Squaring both sides we get,

a^{2}(1 + r + r^{2})^{2} = (21)^{2}….(1)

And from the second condition,

a^{2} + a^{2}r^{2} + a^{2}r^{4} = 189

a^{2}(1 + r^{2} + r^{4}) = 189……(2)

Dividing both the equations we get,

Cross multiplying we get,

3 + 3r + 3r^{2} = 7r^{2} – 7r + 7

4r^{2} – 10r + 4 = 0

2r^{2} – 5r + 2 = 0

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

2r^{2} – (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,

a^{2}(1 + r^{2} + r^{4}) = 189

a^{2}(1 + 4 + 16) = 189

a^{2}

a^{2} = 9

a = ±3

At r = 1/2

a^{2} = 9 × 16

a = 3 × 4 = 12

The numbers are:

1, 9, 81 or 81, 9, 1

Rate this question :

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