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.

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)

Also, (a − 1), (ar − 7), (ar² − 21) form an A.P.

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

⇒ ar − a − 6 = ar² − ar − 14

⇒ ar² − 2ar + a = 8

⇒ a(r² + 1 − 2r) = 8

⇒ a(r − 1)² = 8     ...(2)

From (1) and (2):

7(r² − 2r + 1) = 1 + r + r²

⇒ 7r² − 14r + 7 = r² + r + 1

⇒ 6r² − 15r + 6 = 0

⇒ (6r − 3)(r − 2) = 0

Hence, r = 2 or r = ½.

When r = 2, a = 8 ⇒ numbers are 8, 16, 32.
When r = ½, a = 32 ⇒ numbers are 32, 16, 8.

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