Correct Answer - 8,16,32
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.
