Given the first three terms of a geometric progression are 144, x, and 64.

Let the common ratio be r. Then:

\(x = 144r\)

\(64 = xr = 144r^2\)

Solving for r:

\(r^2 = \frac{64}{144} = \frac{4}{9}\)

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

Substitute r back to find x:

\(x = 144 \times \frac{2}{3} = 96\)

For the sum to infinity:

The formula for the sum to infinity of a geometric progression is:

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

where a is the first term and r is the common ratio.

Substitute the values:

\(S_\infty = \frac{144}{1 - \frac{2}{3}} = \frac{144}{\frac{1}{3}} = 432\)