This geometric sequence can be written as $a+ar+ar^2+ar^3+\cdots$. We are given that $a+ar=4\frac{1}{2}$. Using the [formula](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence#Infinite_Geometric_Sequences) 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 $\boxed{\textbf{(E)}\ 9 \text{ or }3}$.
The sum to infinity of the terms of an infinite geometric progression is $6$. The sum of the first two terms is $4\frac{1}{2}$. The first term of the progression is:
$\textbf{(A) \ }3 \text{ or } 1\frac{1}{2}  \qquad \textbf{(B) \ }1 \qquad \textbf{(C) \ }2\frac{1}{2} \qquad \textbf{(D) \ }6 \qquad \textbf{(E) \ }9\text{ or }3$