Total ways to choose 3 tickets = \(\binom{2n+1}{3} = \frac{(2n+1)(2n)(2n-1)}{6}\).

Three numbers \(a < b < c\) are in AP when \(a + c = 2b\), that is, \(a\) and \(c\) have the same parity (then \(b\) is fixed).

Among 1 to \(2n+1\), there are \(n+1\) odd and \(n\) even numbers.

Favourable = \(\binom{n+1}{2} + \binom{n}{2} = \frac{(n+1)n + n(n-1)}{2} = n^2\)

Probability = \(\frac{6n^2}{(2n+1)(2n)(2n-1)} = \frac{3n}{4n^2 - 1}\)

Answer: (D)