Let the three numbers in G.P. be $a,ar,ar^2$
It given that $a+ar+ar^2=56$......(i)
It is also given that $(a-1),\:\:(ar-7)\:\:and\:\:(ar^2-21)$ are in A.P.
We know that if any three numbers $x,y,z$ are in A.P., then $2y=x+z$
$\Rightarrow\:2(ar-7)=(a-1)+(ar^2-21)$
$\Rightarrow\:2ar-14=a+ar^2-22$
From (i) we get $ a+ar^2=56-ar$
$\therefore\:2ar=(56-ar)-22+14$
$\Rightarrow\:3ar=48$
$\Rightarrow\:ar=16$ $\Rightarrow\:a=\large\frac{16}{r}$
Substituting the value of $ar$ and $a$ in (i) we get
$a+16+ar^2=56$
$\Rightarrow\:a(1+r^2)=40$
$\Rightarrow\:\large\frac{16}{r}$$(1+r^2)=40$
$\Rightarrow\:2+2r^2=5r$
$\Rightarrow\:2r^2-5r+2=0$
$\Rightarrow\:2r^2-4r-r+2=0$
$\Rightarrow\:2r(r-2)-(r-2)=0$
$\Rightarrow\:r=2\:\:or\:\:r=\large\frac{1}{2}$
Substituting the value of $r=2$, we get $a=\large\frac{16}{2}=8$
and if $r=\large\frac{1}{2}$ we get $a=\large\frac{16}{1/2}$$=32$
$\therefore$ The numbers are $8,16\:\;and\:\:32$