The sum of the terms in an infinite geometric series is 15, and the sum of their squares is 45.  Find the first term.
Let $a$ be the first term, and let $r$ be the common ratio.  Then
\begin{align*}
\frac{a}{1 - r} &= 15, \\
\frac{a^2}{1 - r^2} &= 45.
\end{align*}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 $a = 15 \left( 1 - \frac{2}{3} \right) = \boxed{5}.$