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Let the three numbers be

∴ According to the question

⇒ …(1)

⇒ a + ar + ar² = 56r

⇒

⇒ …(2)

Subtracting 1,7,21 we get,

⇒

The above numbers are in AP

If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c

∴

⇒ 2ar – 14r = a – r + ar² – 21r

⇒ ar² – 8r + a – 2ar = 0

⇒ a(r² – 2r + 1) = 8r

From (2) we know the value of a

⇒

⇒ 56(r² – 2r + 1) = 8(1 + r + r²)

⇒ 7(r² – 2r + 1) = (1 + r + r²)

⇒ 7r² – 14r + 7 = 1 + r + r²

⇒ 6r² – 15r + 6 = 0

⇒ 6r² – 12r – 3r + 6 = 0

⇒ 6r(r – 2) – 3(r – 2) = 0

⇒ r = 2 or r = 3/6 = 1/2

When r = 2 ⇒ a = 16 {using equation 1)

r = 1/2 ⇒ a = 16

∴ the three numbers are (a/r, a, ar) = (8,16,32)

Or numbers are – (32,16,8)