Given a geometric progression with first term \(a = 4x\) and second term \(ar = x^2\), we have:

\(r = \frac{x^2}{4x} = \frac{x}{4}\).

The sum to infinity of a geometric series is given by \(\frac{a}{1-r} = 8\).

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

\(\frac{4x}{1 - \frac{x}{4}} = 8\).

Solving for \(x\):

\(\frac{4x}{\frac{4-x}{4}} = 8\)

\(4x \cdot \frac{4}{4-x} = 8\)

\(\frac{16x}{4-x} = 8\)

\(16x = 8(4-x)\)

\(16x = 32 - 8x\)

\(24x = 32\)

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

Thus, \(r = \frac{x}{4} = \frac{1}{3}\).

The third term is \(ar^2 = 4x \left(\frac{1}{3}\right)^2\).

\(ar^2 = 4 \cdot \frac{4}{3} \cdot \frac{1}{9} = \frac{16}{27}\).