(i) For a geometric progression, the ratio between consecutive terms is constant. Therefore,

\(\frac{k + 6}{2k + 3} = \frac{k}{k + 6}\)

Cross-multiplying gives:

\((k + 6)^2 = k(2k + 3)\)

Expanding both sides:

\(k^2 + 12k + 36 = 2k^2 + 3k\)

Rearranging gives:

\(0 = 2k^2 + 3k - k^2 - 12k - 36\)

\(0 = k^2 - 9k - 36\)

Solving the quadratic equation:

\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = -9\), \(c = -36\):

\(k = \frac{9 \pm \sqrt{81 + 144}}{2}\)

\(k = \frac{9 \pm 15}{2}\)

\(k = 12\) or \(k = -3\). Since all terms are positive, \(k = 12\).

(ii) The first term \(a = 2k + 3 = 27\) when \(k = 12\). The common ratio \(r = \frac{k + 6}{2k + 3} = \frac{18}{27} = \frac{2}{3}\).

The sum to infinity \(S_\infty\) of a geometric progression is given by:

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

\(S_\infty = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81\)