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

Explanation

Let the three number in G.P. be `a,ar,ar^(2)`
Sum of numbers is a`(1+r+r^(2))`=56
Subtracting 1,7,21 from these terms, we get terms
`a-1,ar-7,ar^(2)-21`
Now these numbers are in A.P. Thus.
`2(ar-7)=a-1+ar^(2)-21`
or `a(r^(2)-2r+1)=8` (2)
From (1) and (2), we get
`7(r^(2)-2r+1)=1+r+r^(2)`
or `6r^(2)-15r+6=0`
or (6r-3)(r-2)=0
or r=2,`1/2`
When r=2,a=8
when `r=1/2,a=32`
Therefore ,when r=2, the three numbers in G.P, are 8,16, and 32.
When `r=1/2`, the three numbers in G.P are 32,16 and 8.
Thus, in either case, the three required numbers are 8,16 and 32.