If the sum of three numbers in G.P. is 63 and the product of the first and the second term is $\frac{3}{4}$of the third term, then the numbers are

Let the three numbers be a, ar,ar²

Given $\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2=63$-------------- (1)

and $\mathrm{a}\cdot \mathrm{ar}=\frac{3}{4}\cdot {\mathrm{ar}}^2\text{ or }\mathrm{a}=\frac{3}{4}\mathrm{r}$------------------(2)

putting in  (1)  

 $\begin{array}{r}\frac{3}{4}\mathrm{r}+\frac{3}{4}\mathrm{r}\cdot \mathrm{r}+\frac{3}{4}\mathrm{r}\cdot {\mathrm{r}}^2=63\\ \mathrm{or} {\mathrm{r}}^3+{\mathrm{r}}^2+\mathrm{r}-84=0\end{array}$

$\begin{array}{r}\text{ or }(\mathrm{r}-4)\left({\mathrm{r}}^2+5\mathrm{r}+21\right)=0\\ \therefore \mathrm{r}=4,\frac{-5\pm \sqrt{25-84}}{2}\end{array}$

$\therefore$ Real value of r is 4. Putting this value in (2),

   $\mathrm{a}=\frac{3}{4}\times 4=3$

$\therefore$The three numbers are $3,3\times 4,3\times 4^2$,i e  3, 12, 48.