The geometric series $a+ar+ar^2\ldots$ has a sum of $7$, and the terms involving odd powers of $r$ have a sum of $3$. What is $a+r$?
$\textbf{(A)}\ \frac{4}{3}\qquad\textbf{(B)}\ \frac{12}{7}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{7}{3}\qquad\textbf{(E)}\ \frac{5}{2}$
The sum of an infinite geometric series is given by $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.
In this series, $\frac{a}{1-r} = 7$
The series with odd powers of $r$ is given as \[ar + ar^3 + ar^5 ...\]
Its sum can be given by $\frac{ar}{1-r^2} = 3$
Doing a little algebra
$ar = 3(1-r)(1+r)$
$ar = 3\left(\frac{a}{7}\right)(1+r)$
$\frac{7}{3}r = 1 + r$
$r = \frac{3}{4}$
$a = 7(1-r) = \frac{7}{4}$
$a + r =\boxed{ \frac{5}{2}} \Rightarrow \mathrm{(E)}$