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. $\frac{n(n-1)}{2}$ if $n$ is even
• C. $\frac{(n-1{)}^2}{4}$ if $n$ is odd
• D. none of these

Answer 1

Answer: (B), (C)

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

If $a,b,c$ are in $A.P.$, then $a$ and $c$ both are odd or both are even.
Case1: $n$ is even
The num,ber of ways of selection of two even numbers $a$ and $c$ is ${}^{n/2}{C}_2$. Number of ways of selection of two odd numbers is ${}^{n/2}{C}_2$. Hence the number of $A.P.$ is
$2^{n/2}{C}_2=2\frac{\frac{n}{2}(\frac{n}{2}-1)}{2}=\frac{n(n-2)}{4}$
Case 2: n is odd
The number of ways of selection of two odd numbers $a$ and $c$ is ${}^{(n+1)/2}{C}_2$. Hence, the number of $A.P{.}^{\prime }s$ is
${}^{(n+1)/2}{C}_2+{}^{(n-1)/2}{C}_2$=\dfrac{\left(\dfrac{n+1}{2}\right)\left(\dfrac {n+1}2-1\right)}{2}+\dfrac{\left(\dfrac{n-1}{2}\right)\left(\dfrac {n-1}2-1\right)}{2}=\dfrac18(n-1)((n+1)+(n-3))=\dfrac{(n-1)^2}4$