Let the first term be \(a\) and the common ratio be \(r\).

The third term is given by \(ar^2 = 20\).

The sum to infinity is \(\frac{a}{1-r} = 3a\).

From the sum to infinity equation, \(\frac{a}{1-r} = 3a\), we get:

\(\frac{1}{1-r} = 3\)

\(1 = 3(1-r)\)

\(1 = 3 - 3r\)

\(3r = 2\)

\(r = \frac{2}{3}\)

Substitute \(r = \frac{2}{3}\) into \(ar^2 = 20\):

\(a \left(\frac{2}{3}\right)^2 = 20\)

\(a \cdot \frac{4}{9} = 20\)

\(a = 20 \times \frac{9}{4}\)

\(a = 45\)