The sum of three numbers in $G.P.$ is $56.$ If we subtract $1,7,21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

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

From the given condition,

$a+a r+a r^{2}=56$

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

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

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

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

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

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

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

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

From $(1)$ and $(2),$ we get

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

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

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

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

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

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

When $r=2, a=8$

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 $32$