(a) For an arithmetic progression, the difference between consecutive terms is constant. Therefore, we have:

\(6k - k = k + 6 - 6k\)

\(5k = k + 6 - 6k\)

\(5k = k + 6 - 6k\)

\(5k + 6k = k + 6\)

\(11k = k + 6\)

\(10k = 6\)

\(k = \frac{6}{10} = 0.6\)

(b) The common difference \(d\) is \(6k - k = 5k\). Substituting \(k = 0.6\), we get \(d = 3\).

The sum of the first 30 terms \(S_{30}\) is given by:

\(S_{30} = \frac{30}{2} (2 \times 0.6 + 29 \times 3)\)

\(S_{30} = 15 (1.2 + 87)\)

\(S_{30} = 15 \times 88.2\)

\(S_{30} = 1323\)