The sum of 3 numbers in GP is 56. If 1, 7 and 21 are subtracted from them respectively, they form an AP. What are the numbers?
Answers
let the three numbers in G.P. be a, ar, and ar2.
From the given condition, a + ar + ar2 = 56
⇒ a (1 + r + r2) = 56
a – 1, ar – 7, ar2 – 21 forms an A.P.
∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)
⇒ ar – a – 6 = ar2 – ar – 14
⇒ar2 – 2ar + a = 8
⇒ar2 – ar – ar + a = 8
⇒a(r2 + 1 – 2r) = 8
⇒ a (r – 1)2 = 8 … (2)
⇒7(r2 – 2r + 1) = 1 + r + r2
⇒7r2 – 14 r + 7 – 1 – r – r2 = 0
⇒ 6r2 – 15r + 6 = 0
⇒ 6r2 – 12r – 3r + 6 = 0
⇒ 6r (r – 2) – 3 (r – 2) = 0
⇒ (6r – 3) (r – 2) = 0
When r = 2, a = 8
When
Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When , the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.
Answer
Let the three numbers in G.P. be a, ar, and ar2.
From the given condition, a + ar + ar2 = 56
⇒ a (1 + r + r2) = 56

a – 1, ar – 7, ar2 – 21 forms an A.P.
∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)
⇒ ar – a – 6 = ar2 – ar – 14
⇒ar2 – 2ar + a = 8
⇒ar2 – ar – ar + a = 8
⇒a(r2 + 1 – 2r) = 8
⇒ a (r – 1)2 = 8 … (2)

⇒7(r2 – 2r + 1) = 1 + r + r2
⇒7r2 – 14 r + 7 – 1 – r – r2 = 0
⇒ 6r2 – 15r + 6 = 0
⇒ 6r2 – 12r – 3r + 6 = 0
⇒ 6r (r – 2) – 3 (r – 2) = 0
⇒ (6r – 3) (r – 2) = 0

When r = 2, a = 8
When 
Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.
When , the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.