The sum of three numbers in G.P. is 66. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Let the three numbers in G.P. be a, ar, and ar².

According to question -

a + ar + ar² = 66

⇒ a(1 + r + r²) = 66

…(1)

Given that -

(a – 1), (ar – 7), (ar² – 21) forms an A.P.

thus the common difference between the consecutive terms will be equal.

∴ (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)

⇒ ar – a – 6 = ar² – ar – 14

⇒ ar² – 2ar + a = 8

⇒ a(r² – 2r + 1) = 8

…(2)

Comparing equations (1) & (2), we get -

On Cross - multiplying,

⇒66(r² – 2r + 1) = 8(1 + r + r²)

⇒7(r² – 2r + 1) = (1 + r + r²)

⇒ 7r² – 14r + 7 = 1 + r + r²

⇒ 6r² – 16r + 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

When r = 2, a = 8

When r = 1/2, a = 32

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.