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

The second term is \(ar = 24\).

The fourth term is \(ar^3 = 13\frac{1}{2} = \frac{27}{2}\).

From \(ar = 24\), we have \(a = \frac{24}{r}\).

Substitute \(a\) in \(ar^3 = \frac{27}{2}\):

\(\frac{24}{r} \cdot r^3 = \frac{27}{2}\)

\(24r^2 = \frac{27}{2}\)

\(r^2 = \frac{27}{48} = \frac{9}{16}\)

\(r = \frac{3}{4}\) (since \(r\) is positive)

Substitute \(r\) back to find \(a\):

\(a = \frac{24}{\frac{3}{4}} = 32\)

Thus, the first term \(a = 32\).

For the sum to infinity:

The formula for the sum to infinity is \(S_\infty = \frac{a}{1-r}\).

\(S_\infty = \frac{32}{1 - \frac{3}{4}} = \frac{32}{\frac{1}{4}} = 128\).