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

The second term is \(ar = 10\) and the third term is \(ar^2 = 8\).

From \(ar^2 = 8\) and \(ar = 10\), we have:

\(\frac{ar^2}{ar} = \frac{8}{10}\)

\(r = 0.8\)

Substitute \(r = 0.8\) into \(ar = 10\):

\(a \times 0.8 = 10\)

\(a = \frac{10}{0.8} = 12.5\)

The sum to infinity \(S_\infty\) is given by:

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

\(S_\infty = \frac{12.5}{1-0.8}\)

\(S_\infty = \frac{12.5}{0.2}\)

\(S_\infty = 62.5\)