The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.
Let the three numbers be
∴ According to the question
⇒ …(1)
⇒ a + ar + ar2 = 56r
⇒
⇒ …(2)
Subtracting 1,7,21 we get,
⇒
The above numbers are in AP
If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c
∴
⇒ 2ar – 14r = a – r + ar2 – 21r
⇒ ar2 – 8r + a – 2ar = 0
⇒ a(r2 – 2r + 1) = 8r
From (2) we know the value of a
⇒
⇒ 56(r2 – 2r + 1) = 8(1 + r + r2)
⇒ 7(r2 – 2r + 1) = (1 + r + r2)
⇒ 7r2 – 14r + 7 = 1 + r + r2
⇒ 6r2 – 15r + 6 = 0
⇒ 6r2 – 12r – 3r + 6 = 0
⇒ 6r(r – 2) – 3(r – 2) = 0
⇒ r = 2 or r = 3/6 = 1/2
When r = 2 ⇒ a = 16 {using equation 1)
r = 1/2 ⇒ a = 16
∴ the three numbers are (a/r, a, ar) = (8,16,32)
Or numbers are – (32,16,8)