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

For progression P, the first term \(a = 6\) and the second term is \(6r\), so the common ratio \(r\) is \(r\).

The sum to infinity for P is \(S_P = \frac{6}{1 - r}\).

For progression Q, the first term \(a = 12\) and the second term is \(-12r\), so the common ratio \(r = -r\).

The sum to infinity for Q is \(S_Q = \frac{12}{1 + r}\).

Since the sums are equal, \(\frac{6}{1 - r} = \frac{12}{1 + r}\).

Cross-multiplying gives:

\(6(1 + r) = 12(1 - r)\)

\(6 + 6r = 12 - 12r\)

\(18r = 6\)

\(r = \frac{1}{3}\)

Substitute \(r = \frac{1}{3}\) back into the sum formula for P:

\(S = \frac{6}{1 - \frac{1}{3}} = \frac{6}{\frac{2}{3}} = 9\)