Question

Number of ways in which three numbers in $A.P.$ can be selected from $\{1,2,3,.....,n\}$ is

• A. ${\left(\frac{n-1}{2}\right)}^2$ if $n$ is even
  
• B. $\frac{n(n-2)}{4}$ if $n$ is even
  
• C. $\frac{(n-1{)}^2}{4}$ if $n$ is odd
  
• D. $\frac{n(n-2)}{4}$ if $n$ is odd

Answer 1

Answer: (B), (C)

$\frac{(n-1{)}^2}{4}$ if $n$ is odd


If $a,b,c$ are in $A.P.$, then $a+c=2b$
$\Rightarrow$$a$ and $c$ both are odd or both are even.
Case $I$: If $n$ is even.
The number of ways of selection of two even numbers $a$ and $c$ is ${}^{\frac{n}{2}}{C}_2$ and Number of ways of selection of two odd numbers is ${}^{\frac{n}{2}}{C}_2.$
Hence the number of $A.P^{\prime }s$ possible are
$2\times {}^{\frac{n}{2}}{C}_2=2\times \frac{\frac{n}{2}(\frac{n}{2}-1)}{2}=\frac{n(n-2)}{4}$

Case $II$: If $n$ is odd.
The number of ways of selection of two odd numbers $a$ and $c$ is${}^{\frac{(n+1)}{2}}{C}_2$ and
Number of ways of selection of two even numbers is ${}^{\frac{(n-1)}{2}}{C}_2$

Hence the number of $A.P^{\prime }s$ possible are
$=\frac{\left(\frac{n+1}{2}\right)(\frac{n+1}{2}-1)}{2}+\frac{\left(\frac{n-1}{2}\right)(\frac{n-1}{2}-1)}{2}$
$=\frac{1}{8}\left[\right(n-1\left)\right(n+1)+(n-3\left)\right]$
$=\frac{(n-1{)}^2}{4}$