(i) Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). The second term is given by \(ar = 3\). The sum to infinity of a geometric progression is given by \(\frac{a}{1-r} = 12\).

We have the equations:

\(ar = 3\)

\(\frac{a}{1-r} = 12\)

From \(ar = 3\), we can express \(r\) as \(r = \frac{3}{a}\).

Substitute \(r = \frac{3}{a}\) into \(\frac{a}{1-r} = 12\):

\(\frac{a}{1-\frac{3}{a}} = 12\)

\(\frac{a^2}{a-3} = 12\)

\(a^2 = 12(a-3)\)

\(a^2 = 12a - 36\)

\(a^2 - 12a + 36 = 0\)

Solving this quadratic equation, we find \(a = 6\).

(ii) For the arithmetic progression, the first term \(a = 6\) and the second term is 3, so the common difference \(d = 3 - 6 = -3\).

The sum of the first 20 terms of an arithmetic progression is given by:

\(S_{20} = \frac{n}{2} (2a + (n-1)d)\)

\(S_{20} = \frac{20}{2} (2 \times 6 + 19 \times (-3))\)

\(S_{20} = 10 (12 - 57)\)

\(S_{20} = 10 \times (-45)\)

\(S_{20} = -450\)