Let the first term of both progressions be 3. For the arithmetic progression (AP), let the common difference be \(d\). The terms are:

1st term: \(3\)

3rd term: \(3 + 2d\)

13th term: \(3 + 12d\)

For the geometric progression (GP), let the common ratio be \(r\). The terms are:

1st term: \(3\)

2nd term: \(3r\)

3rd term: \(3r^2\)

Equating the terms of the AP and GP:

\(3 + 2d = 3r\)

\(3 + 12d = 3r^2\)

From \(3 + 2d = 3r\), we get \(r = \frac{3 + 2d}{3}\).

Substitute \(r\) in the second equation:

\(3 + 12d = 3\left(\frac{3 + 2d}{3}\right)^2\)

\((3 + 2d)^2 = 3(3 + 12d)\)

\(9 + 12d + 4d^2 = 9 + 36d\)

\(4d^2 - 24d = 0\)

\(4d(d - 6) = 0\)

\(d = 0\) or \(d = 6\)

Since \(d = 0\) is not valid, \(d = 6\).

Substitute \(d = 6\) back to find \(r\):

\(r = \frac{3 + 2(6)}{3} = \frac{15}{3} = 5\)

Thus, the common difference is 6 and the common ratio is 5.