Let the first term be \(a = 16\) and the common ratio be \(r\).

The fourth term is given by \(ar^3 = \frac{27}{4}\).

Substitute \(a = 16\) into the equation:

\(16r^3 = \frac{27}{4}\)

Divide both sides by 16:

\(r^3 = \frac{27}{64}\)

Take the cube root of both sides:

\(r = \frac{3}{4}\)

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

\(S = \frac{a}{1 - r}\)

Substitute \(a = 16\) and \(r = \frac{3}{4}\):

\(S = \frac{16}{1 - \frac{3}{4}}\)

\(S = \frac{16}{\frac{1}{4}}\)

\(S = 16 \times 4 = 64\)