Let the first term of the geometric progression be \(a\) and the common ratio be \(r\).

The second term is \(ar\).

The sum to infinity \(S\) of a geometric progression is given by \(S = \frac{a}{1-r}\), provided \(|r| < 1\).

According to the problem, \(ar = 0.24 \times \frac{a}{1-r}\).

Thus, we have:

\(ar = \frac{24}{100} \times \frac{a}{1-r}\)

\(ar = \frac{24a}{100(1-r)}\)

\(100ar(1-r) = 24a\)

\(100ar - 100ar^2 = 24a\)

Divide through by \(a\) (assuming \(a \neq 0\)):

\(100r - 100r^2 = 24\)

\(100r^2 - 100r + 24 = 0\)

Factor the quadratic equation:

\((20r - 8)(5r - 3) = 0\)

Setting each factor to zero gives:

\(20r - 8 = 0 \quad \Rightarrow \quad r = \frac{8}{20} = \frac{2}{5}\)

\(5r - 3 = 0 \quad \Rightarrow \quad r = \frac{3}{5}\)

Thus, the possible values of the common ratio are \(r = \frac{2}{5}\) and \(r = \frac{3}{5}\).