Solution:

Suppose that the three numbers are a, ar, ar²

According to the question statement,

a + ar + ar² = 56 … (1)

subtracting 1,7,21 respectively from these terms,we get:

(a – 1), (ar – 7), (ar² – 21)

These terms are in AP and we know that If three numbers are in AP, we can write 2b = a + c. Therefore,

$ 2\text{ }\left( ar-7 \right)\text{ }=\text{ }a-1\text{ }+\text{ }a{{r}^{2}}-21 $

$ =\text{ }\left( a{{r}^{2}}~+\text{ }a \right)-22 $

$ 2ar-14\text{ }=\text{ }\left( 56-ar \right)-22 $

$ 2ar-14\text{ }=\text{ }34-ar $

$ 3ar\text{ }=\text{ }48 $

$ ar\text{ }=\text{ }48/3 $

$ ar\text{ }=\text{ }16 $

$ a\text{ }=\text{ }16/r\text{ }\ldots .\text{ }\left( 2 \right) $

Upon putting the value of a in equation (1) we get,

$ \left( 16\text{ }+\text{ }16r\text{ }+\text{ }16{{r}^{2}} \right)/r\text{ }=\text{ }56 $

$ 16\text{ }+\text{ }16r\text{ }+\text{ }16{{r}^{2}}~=\text{ }56r $

$ 16{{r}^{2}}-40r\text{ }+\text{ }16\text{ }=\text{ }0 $

$ 2{{r}^{2}}-5r\text{ }+\text{ }2\text{ }=\text{ }0 $

$ 2{{r}^{2}}-4r-r\text{ }+\text{ }2\text{ }=\text{ }0 $

$ 2r\left( r-2 \right)-1\left( r-2 \right)\text{ }=\text{ }0 $

$ \left( r-2 \right)\text{ }\left( 2r-1 \right)\text{ }=\text{ }0 $

$ r\text{ }=\text{ }2\text{ }or\text{ }1/2 $

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

a = 16/r = 16/2 or 16/(1/2)

a = 8 or 32

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