Thank you for reporting, we will resolve it shortly
Q.
The number of ways in which three numbers belongs to A.P. can be selected from $1, 2, 3, ...n$ if $n = 2m$, is
Permutations and Combinations
Solution:
Here, $n = 2m$ means $n$ is even. In this case, the max. value of the common difference $d = m - 1$ and there are two $A.P.’s$ corresponding to the maximum $d$
i.e., $( 1, m, 2m - 1)$ and $(2, m + 1, 2m)$.
The A.P.’s corresponding to other values of $d$ are given as
$\therefore $ Number of $A.P.'s$
$= 2 + 4 + 6 + ... + (2m - 4) + (2m - 2)$
$= 2 [1 + 2 + 3 + ... + (m -1)]$
$ = (m - 1)(m) = \frac{n(n-2)}{4} $
$\left[\because n = 2m \therefore m = \frac{n}{2}\right]$
