smc

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 $$ar+a{r}^3+a{r}^5...$$ Its sum can be given by $\frac{ar}{1-r^2}=3$ Doing a little algebra $ar=3(1-r)(1+r)$ $ar=3\left(\frac{a}{7}\right)(1+r)$ $\frac{7}{3}r=1+r$ $r=\frac{3}{4}$ $a=7(1-r)=\frac{7}{4}$ $a+r=\boxed{\frac{5}{2}}\Rightarrow (\mathrm{E})$