Arithmetic probability

Let the $3$ numbers chosen from the $2n + 1$ consecutive integers be arranged in ascending order. Then if they are in AP, there will be a minimum difference (magnitude) of $d = 1$ between successive terms, and a maximum difference (magnitude) of $d = n,$ (since $2n + 1$ is always odd).

There will be $(2n + 1) - 2$ AP sequences with $d = 1$ , $(2n + 1) - 4$ AP sequences with $d = 2,$ and in general $(2n + 1) - 2m$ AP sequences with $d = m.$ Thus the number of AP's that can be obtained is

$\displaystyle\sum_{k=1}^{n} (2k - 1) = 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = n(n + 1) - n = n^{2}.$

Now there are $\dbinom{2n + 1}{3} = \dfrac{(2n + 1)(2n)(2n - 1)}{6} = \dfrac{n(4n^{2} - 1)}{3}$ possible draws without restrictions, so the desired probability is

$\dfrac{n^{2}}{\dfrac{n(4n^{2} - 1)}{3}} = \boxed{\dfrac{3n}{4n^{2} - 1}}.$

In any 3 term AP, the first and third term must have the same parity (odd or even) If They do, there will be an integral second term which is the mean of the first and third. So, number of desired outcomes is equal to the number of ways of picking 2 numbers from all odd terms + number of ways of picking two numbers out of all even terms.

This would be Nc2 + N+1c2 which equals n^2

Lol,put n=1 .!!!! . . . . . . . . . . .