G.P. is increasing, i.e. $r>1.$
Given, $a+a{r}^{n-1}=99, ar\cdot a{r}^{n-2}=288$ and $\frac{a\left(1-r^n\right)}{1-r}=189$ .
$a\left(1+r^{n-1}\right)=99$ and $a^2{r}^{n-1}=288$
$\Rightarrow a\left(1+\frac{288}{a^2}\right)=99$
$\Rightarrow a^2+288=99a\Rightarrow a^2-99a+288=0$
$\Rightarrow a=\mathrm{3,} 96\Rightarrow r^{n-1}=\frac{288}{a^2}=32, \frac{1}{32}$
As $r>1\Rightarrow r^{n-1}=32 \mathrm{for} a=3$
Now, $\frac{a\left(1-r^n\right)}{1-r}=189\Rightarrow \frac{3\left(1-r\cdot r^{n-1}\right)}{1-r}=189$
$\Rightarrow \left(\frac{1-r\cdot 32}{1-r}\right)=63\Rightarrow 1-32r=63-63r$
$\Rightarrow 31r=62\Rightarrow r=2.$

Now, $2^{n-1}=32\Rightarrow n=6.$