jmc

Let $a$ be the first term, and let $r$ be the common ratio. Then $$\begin{array}{rl}\frac{a}{1-r}& =15,\\ \frac{a^2}{1-r^2}& =45.\end{array}$$From the first equation, $a=15(1-r).$ Substituting into the second equation, we get $$\frac{225(1-r{)}^2}{1-r^2}=45.$$The denominator factors as $(1+r)(1-r),$ so the equation simplifies to $$\frac{5(1-r)}{1+r}=1.$$Then $5-5r=1+r,$ so $r=\frac{2}{3}.$ Then $a=15\left(1-\frac{2}{3}\right)=\boxed{5}.$