Find three numbers in G.P. whose sum is 38 and their product is 1728.

Let the terms of the the given G.P. be $\frac{a}{r}, a$ and $a r$.

Then, product of the G.P. = 1728

= a3 = 1728

= a = 12

Similarly, sum of the G.P. = 38

$\Rightarrow \frac{a}{r}+a+a r=38$

Substituting the value of a

$\frac{12}{r}+12+12 r=38$

$\Rightarrow 12 r^{2}+12 r+12=38 r$

$\Rightarrow 12 r^{2}-26 r+12=0$

$\Rightarrow 2\left(6 r^{2}-13 r+6\right)=0$

$\Rightarrow 6 r^{2}-13 r+6=0$

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

$\Rightarrow r=\frac{2}{3}, \frac{3}{2}$

Hence, the G.P. for $a=12$ and $r=\frac{2}{3}$ is 18,12 and 8 .

And, the G.P. for $a=12$ and $r=\frac{3}{2}$ is 8,12 and 18 .

Hence, the three numbers are 8, 12 and 18.