Question

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

Answer 1

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

From the given condition

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

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

$\therefore (ar-7)-(a-1)=({ar}^2-21)-(ar-7)\Rightarrow ar-a-6={ar}^2-ar-14\Rightarrow {ar}^2-2ar+a=8\Rightarrow a(r^2+1-2r)=8\Rightarrow a{(r-1)}^2=8...\left(2\right)$

Using (1) and (2)

$\Rightarrow 7(r^2-2r+1)=1+r+r^2\Rightarrow {7r}^2-14r+7-1-r-r^2=0\Rightarrow {6r}^2-15r+6=0\Rightarrow {6r}^2-12r-3r+6=0\Rightarrow 6r(r-2)-3(r-2)=0\Rightarrow (6r-3)(r-2)=0\Rightarrow r=2,\frac{1}{2}$

When $r=2$, $a=8$

And three numbers in GP are $8,16$ $32$.

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