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

Solution:

Let the three numbers in G.P. be $a, a r$, and $a r^{2}$.

From the given condition, $a+a r+a r^{2}=56$

$\Rightarrow a\left(1+r+r^{2}\right)=56$

$\Rightarrow a=\frac{56}{1+r+r^{2}}$  (1)

$a-1, a r-7, a r^{2}-21$ forms an A.P.

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

$\Rightarrow a r-a-6=a r^{2}-a r-14$

$\Rightarrow a r^{2}-2 a r+a=8$

$\Rightarrow a r^{2}-a r-a r+a=8$

$\Rightarrow a\left(r^{2}+1-2 r\right)=8$

$\Rightarrow a(r-1)^{2}=8 \ldots(2)$

$\Rightarrow \frac{56}{1+r+r^{2}}(r-1)^{2}=8$ [Using (1)]

$\Rightarrow 7\left(r^{2}-2 r+1\right)=1+r+r^{2}$

$\Rightarrow 7 r^{2}-14 r+7-1-r-r^{2}=0$

$\Rightarrow 6 r^{2}-15 r+6=0$

$\Rightarrow 6 r^{2}-12 r-3 r+6=0$

$\Rightarrow 6 r(r-2)-3(r-2)=0$

$\Rightarrow(6 r-3)(r-2)=0$

$\therefore r=2, \frac{1}{2}$

When $r=2, a=8$

When $r=\frac{1}{2}, a=32$

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

When $r=\frac{1}{2}$, the three numbers in G.P. are 32, 16, and 8 .

Thus, in either case, the three required numbers are 8, 16, and 32.