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 first term of a G.P be a and its common ratio be r.

$\therefore a_{1}+a_{2}+a_{3}=56$

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

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

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

Now, according to the question:

$a-1, a r-7$ and $a r^{2}-21$ are in A.P.

$\therefore 2(\operatorname{ar}-7)=a-1+a r^{2}-21$

$\Rightarrow 2 \mathrm{ar}-14=\mathrm{ar}^{2}+\mathrm{a}-22$

$\Rightarrow \mathrm{ar}^{2}-2 \mathrm{ar}+\mathrm{a}-8=0$

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

$\Rightarrow \mathbf{a}=\frac{8}{(1-\mathrm{r})^{2}}$    ...(ii)

Equating $(\mathrm{i})$ and $(\mathrm{ii})$ :

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

$\Rightarrow 8\left(1+\mathrm{r}+\mathrm{r}^{2}\right)=56\left(1+\mathrm{r}^{2}-2 \mathrm{r}\right) \Rightarrow 1+\mathrm{r}+\mathrm{r}^{2}=7\left(1+\mathrm{r}^{2}-2 \mathrm{r}\right)$

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

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

$\Rightarrow 3\left(2 \mathrm{r}^{2}-5 \mathrm{r}+2\right)=0$

 

$\Rightarrow 2 \mathrm{r}^{2}-4 \mathrm{r}-\mathrm{r}+2=0$

$\Rightarrow 2 \mathrm{r}(\mathrm{r}-2)-1(\mathrm{r}-2)=0$

$\Rightarrow(\mathrm{r}-2)(2 \mathrm{r}-1)=0$

$\Rightarrow \mathrm{r}=2, \frac{1}{2}$

When $\mathrm{r}=2, \mathrm{a}=8 . \quad[\operatorname{Using}(\mathrm{ii})]$

And, the required numbers are 8,16 and 32 .

When $\mathrm{r}=\frac{1}{2}, \mathrm{a}=32 . \quad[\mathrm{Using}(\mathrm{ii})]$

And, the required numbers are 32,16 and 8 .