The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.
Let the three numbers be .
∴ According to the question
⇒
⇒ a + ar + ar2 = 14r…(1)
First two terms are increased by 1, and third decreased by 1
∴
The above sequence is in AP.
We know in AP.
2b = a + c
∴
⇒
⇒ 2ar + 2r = ar2 + a
⇒ ar2 – 2ar + a = 2r …(2)
Dividing 1 by 2 we get
⇒
⇒
⇒ 1 + r + r2 = 7r2 – 14r + 7
⇒ 6r2 – 15r – 6 = 0
⇒ 6r2 – 12r – 3r – 6 = 0
⇒ 6r(r – 2) – 3(r – 2) = 0
⇒ (6r – 3) (r – 2) = 0
⇒ r = 2 or r = 1/2.
Substituting r = 2 in 2 we get
⇒ a(2)2 – 2a(2) + a = 2(2)
⇒ 4a – 4a + a = 4
⇒ a = 4
Substituting r = 1/2 in 2 we get
⇒ a(1/2)2 – 2a(1/2) + a = 2(1/2)
⇒ a = 4
∴ substituting a and r we get the numbers as 2,4,8.