Question

The number of ways in which three numbers belongs to A.P. can be selected from $1,2,3,...n$ if $n=2m$, is

• A. $n-3!$
• B. $\frac{n-3}{4}$
• C. $\frac{n(n-2)}{4}$
• D. $\frac{(n-1{)}^2}{4}$

Answer 1

Answer: (C)

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 $A.P{.}^{\prime }s$
$=2+4+6+...+(2m-4)+(2m-2)$
$=2[1+2+3+...+(m-1)]$
$=(m-1)(m)=\frac{n(n-2)}{4}$
$\left[\because n=2m\therefore m=\frac{n}{2}\right]$