Let the first term be \(a = 12\) and the common ratio be \(r\).

The sum to infinity \(S_\infty\) is given by \(\frac{a}{1-r}\).

The sum of the first four terms \(S_4\) is \(\frac{a(1-r^4)}{1-r}\).

According to the problem, \(S_\infty = 9 \times S_4\).

Substitute the formulas: \(\frac{a}{1-r} = 9 \times \frac{a(1-r^4)}{1-r}\).

Cancel \(a\) and \(1-r\) from both sides: \(1 = 9(1-r^4)\).

Simplify: \(9r^4 = 8\).

Thus, \(r^4 = \frac{8}{9}\).

The fifth term is \(ar^4 = 12 \times \frac{8}{9}\).

Calculate: \(ar^4 = \frac{96}{9} \approx 10.7\).