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

The third term is given by \(ar^2 = 48\).

The fourth term is given by \(ar^3 = 32\).

Divide the second equation by the first to find \(r\):

\(\frac{ar^3}{ar^2} = \frac{32}{48}\)

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

Substitute \(r = \frac{2}{3}\) into \(ar^2 = 48\):

\(a \left( \frac{2}{3} \right)^2 = 48\)

\(a \cdot \frac{4}{9} = 48\)

\(a = 108\)

The sum to infinity \(S_\infty\) of a geometric progression is given by:

\(S_\infty = \frac{a}{1 - r}\)

\(S_\infty = \frac{108}{1 - \frac{2}{3}}\)

\(S_\infty = \frac{108}{\frac{1}{3}}\)

\(S_\infty = 324\)