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)