The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

Given \(a = 6\) and \(S_\infty = 18\), we have:

\(\frac{6}{1-r} = 18\)

Solving for \(r\):

\(6 = 18(1-r)\)

\(6 = 18 - 18r\)

\(18r = 12\)

\(r = \frac{2}{3}\)

The new progression is formed by squaring each term of the original progression. The first term of the new progression is \(a^2 = 6^2 = 36\), and the new common ratio is \(r^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

The sum to infinity of the new progression is:

\(S_\infty = \frac{36}{1 - \frac{4}{9}}\)

\(S_\infty = \frac{36}{\frac{5}{9}}\)

\(S_\infty = 36 \times \frac{9}{5}\)

\(S_\infty = \frac{324}{5}\)