The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from t

1.	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 arithmetic progression. Find the numbers.
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 ……………………….… (1) 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) From (1) and (2), we get ⇒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, r=1/2, the three numbers in G.P. are 32, 16, and 8. Thus, in either case, the three required numbers are 8, 16, and 32.

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 arithmetic progression. Find the numbers.

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 ……………………….… (1)

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)

From (1) and (2), we get

⇒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, r=1/2, the three numbers in G.P. are 32, 16, and 8.

Thus, in either case, the three required numbers are 8, 16, and 32.