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.