Question

The number of ways in which three numbers belongs to AP 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)

$\frac{n(n-2)}{4}$

Given numbers are $1,2,3,...n.$
Let the three selected numbers in $A.P.$ be $a,b,c,$ then
$b=\frac{a+c}{2}$ or $a+c=2b$ ...(1)
From (1) it is clear that $a+c$ should be an even integer.
This is possible only when both $a$ and $C$ are odd or both are even.
Case I. When $n$ is even. Let $n=2m$
The number of even number $=m$
and number of even number $=m$
$\therefore$ number of selections of $a$ and $c$ odd number ${=}^m{C}_2$
number of selections of $a$ and $c$ from $m$ even numbers ${=}^m{C}_2$
$\therefore$ number of ways in this case $={2.}^m{C}_2=m(m-1)=\frac{n}{2}(\frac{n}{2}-1)=\frac{n(n-2)}{4}.$