Let the first term be denoted by \(a = 50\) and the common ratio by \(r\).

The third term of the geometric progression is given by \(ar^2 = 32\).

Substituting the value of \(a\), we have:

\(50r^2 = 32\)

Solving for \(r^2\), we get:

\(r^2 = \frac{32}{50} = \frac{16}{25}\)

Thus, \(r = \frac{4}{5}\) (since all terms are positive).

The sum to infinity of a geometric progression is given by:

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

Substituting the values of \(a\) and \(r\), we have:

\(S_\infty = \frac{50}{1 - \frac{4}{5}} = \frac{50}{\frac{1}{5}} = 250\)