Question

Given that n is odd, the number of ways in which three numbers in AP can be selected from $1,2,3,....n$ is

• A. $\frac{(n-1{)}^2}{2}$
• B. $\frac{(n+1{)}^2}{4}$
• C. $\frac{(n+1{)}^2}{2}$
• D. $\frac{(n-1{)}^2}{4}$

Answer 1

The correct option is D $\frac{(n-1{)}^2}{4}$

Let the numbers in A.P. be $a,b,c$
$\Rightarrow a+c=2b$
i.e $a+c$ is an even number which is possible only when a and c both are even or both are odd.
Given n is odd; Let $n=2k+1$
Number of odd numbers from 1 to $n=k+1$
Number of even numbers from 1 to $n=k$
number of ways in which numbers are selected ${=}^{k+1}{C}_2+{}^k{C}_2$
${}^{k+1}{C}_2+{}^k{C}_2=\frac{(k+1)k}{2}+\frac{k(k-1)}{2}=k^2={\left(\frac{n-1}{2}\right)}^2$