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)

$\frac{(n-1{)}^2}{4}$ 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$$=\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}$
$=\frac{1}{8}(n-1)\left(\right(n+1)+(n-3\left)\right)$
$=\frac{(n-1{)}^2}{4}$