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 = 48/3
ar = 16
a = 16/r …. (2)
Now, substitute the value of a in equation (1) we get,
(16 + 16r + 16r2)/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 1/2
Substitute the value of r in equation (2) we get,
a = 16/r
= 16/2 or 16/(1/2)
= 8 or 32
∴ The three numbers are (a, ar, ar2) is (8, 16, 32)