Note that the first series is $\frac{ar(r^{10}-1)}{r-1}=18$ and the second is $\frac{r^{10}-1}{a(r-1)r^{10}}=6$. Now we find that $\frac{18}{a^2r^{11}}=6$, so $a^2r^{11}=3$. Now note that te product of the ten numbers is $a^{10}r^{55}$, so our answer is simply $3^5=243$.