smc

We know that the formula for the sum of an infinite geometric series is $S=\frac{a}{1-r}$. So we can apply this to the conditions given by the problem. We have two equations: $$\begin{array}{rl}15& =\frac{a}{1-r}\\ 45& =\frac{a^2}{1-r^2}\end{array}$$ We get $$\begin{array}{rl}a& =15-15r\\ a^2& =45-45r^2\\ & \\ (15-15r{)}^2& =45-45r^2\\ 270r^2-450r+180& =0\\ 3r^2-5r+2& =0\\ (3r-2)(r-1)& =0\end{array}$$ Since $|r|<1$, $r=\frac{2}{3}$, so plug this into the equation above and we get $a=15-15r=15-10=\boxed{5}$ Solution by $\underline{\text{Invoker}}$