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 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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