jmc

The geometric series involving the odd powers of $r$ is $ar+a{r}^3+a{r}^5+\cdots =5.$ Note that if we subtract this from the original series, the series involving the even powers of $r$ is $$12-5=7=a+a{r}^2+a{r}^4+\cdots =\frac{1}{r}(ar+a{r}^3+a{r}^5+\cdots ).$$However, the series involving even powers of $r$ is just $\frac{1}{r}$ times the series involving odd powers of $r,$ as shown above. Thus, substituting in our values for both those series, $7=\frac{1}{r}(5)⟹r=\boxed{\frac{5}{7}}.$