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. None of these

Answer 1

Answer: (B), (C)

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

If $a,b,c$ are in A.P, then $a$ and $c$ both are odd or both are even.
Case I: $n$ is even
The number 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.'s is
$2^{n/2}{C}_2=2\frac{\frac{n}{2}(\frac{n}{2}-1)}{2}=\frac{n(n-2)}{4}$
Case II: $n$ is odd.
The number of ways of selection of two odd numbers a and c is ${}^{(n+1)/2}{C}_2$. The number of ways of selection of two even numbers a and c is ${}^{(n-1)/2}{C}_2$. Hence, the number of A.P's is
${}^{(n+1)/2}{C}_2{+}^{(n-1)/2}{C}_2$
$=\frac{\left(\frac{n+1}{2}\right)(\frac{n+1}{2}-1)}{2}+\frac{\left(\frac{n-1}{2}\right)(\frac{n-1}{2}-1)}{2}$
$=\frac{1}{8}(n-1)\left(\right(n+1)+(n-3\left)\right)$
$=\frac{(n-1{)}^2}{4}$