Given that $n$is odd, the number of  ways in which three numbers in $AP$ can be selected from $1, 2, 3, ... , n$ is

Let $n=2m+1$Then, the common difference of the

A.P. can be $1, 2, 3, ... , m.$ The number of A.P.'s with $1, 2, 3, ... , m$

common differences are $(2m-1),(2m-3),\dots ,1$ respectively.

$\therefore$Total number of A.P.'s $=(2m-1)+\dots .+1=m^2={\left(\frac{n-1}{2}\right)}^2$

$\therefore$ Total number of ways $={\left(\frac{n-1}{2}\right)}^2$.