smc

This geometric sequence can be written as $a+ar+a{r}^2+a{r}^3+\cdots$. We are given that $a+ar=4\frac{1}{2}$. Using the formula for the sum of an infinite geometric series, we know that $\frac{a}{1-r}=6$. Solving for $r$ in the second equation, we find that $r=\frac{6-a}{6}$. Plugging this into the first equation results in $a^2-12a+27=0$, which can be factored as $(a-3)(a-9)=0$. Hence, $a$ equals $\boxed{9\text{ or }3}$.