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

Let the three numbers be a, ar, ar²

According to the question

a + ar + ar² = 56 … (1)

Let us subtract 1,7,21 we get,

(a – 1), (ar – 7), (ar² – 21)

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

2 (ar – 7) = a – 1 + ar² – 21

= (ar² + a) – 22

2ar – 14 = (56 – ar) – 22

2ar – 14 = 34 – ar

3ar = 48

ar = \( \dfrac{48}{3}\)

ar = 16

a = \( \dfrac{16}{r}\) …. (2)

Now, substitute the value of a in equation (1) we get,

\(\dfrac{ (16 + 16r + 16r^2)}{r}\) = 56

16 + 16r + 16r² = 56r

16r² – 40r + 16 = 0

2r² – 5r + 2 = 0

2r² – 4r – r + 2 = 0

2r(r – 2) – 1(r – 2) = 0

(r – 2) (2r – 1) = 0

r = 2 or \( \dfrac{1}{2}\)

Substitute the value of r in equation (2) we get,

a = \( \dfrac{16}{r}\)

= \( \dfrac{16}{2}\)

= 8 or 32

The three numbers are (a, ar, ar²) is (8, 16, 32)