Question

Find the number of ways in which three numbers in A.P. can be selected from $1,2,3,\dots n.$

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

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{(n-1{)}^2}{4}$ if $n$ is odd
B $\frac{n(n-2)}{4}$ if $n$ is even


If $a,b,c$ are in A.P., then $a$ and $c$ both are even or both are odd.

Case - 1: If $n$ is even.
The number of ways of selecting two even numbers, $a$ and $c$ is ${}^{\frac{n}{2}}{C}_2$. Number of ways of selecting two odd numbers, $a$ and $c$ is ${}^{\frac{n}{2}}{C}_2$.
Hence, number of ways is ${}^{\frac{n}{2}}{C}_2+{}^{\frac{n}{2}}{C}_2=2\times \frac{\frac{n}{2}(\frac{n}{2}-1)}{2}$
$=\frac{n(n-2)}{4}$

Case - 2: If $n$ is odd.
The number of ways of selecting two odd numbers, $a$ and $c$ is ${}^{\frac{(n+1)}{2}}{C}_2$. The number of ways of selecting two even numbers, $a$ and $c$ is ${}^{\frac{(n-1)}{2}}{C}_2$.
Hence, number of ways is ${}^{\frac{(n+1)}{2}}{C}_2+{}^{\frac{(n-1)}{2}}{C}_2$
$=\frac{\left(\frac{n+1}{2}\right)[\left(\frac{n+1}{2}\right)-1]}{2}+\frac{\left(\frac{n-1}{2}\right)[\left(\frac{n-1}{2}\right)-1]}{2}$
$=\frac{1}{8}(n-1)\left[\right(n+1)+(n-3\left)\right]$
$=\frac{(n-1{)}^2}{4}$