Question

The number of ways in which three numbers in A.P. can be seleced from the set of first n natural number if n is odd is

• A. $\frac{n(n-2)}{4}$
• B. $\frac{n{(n-1)}^2}{4}$
• C. $\frac{{(n-1)}^2}{4}$
• D. None of these

Answer 1

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

In order to solve this question, we must observe the number of ways in which we can select the first term of the required A.P. for different values of the common difference($r$) starting from $r=1$ given that there are only $3$ terms before $n$.
For $r=1$, the number of ways in which we can select the first term of the A.P. $=n-2$
For $r=2$, the number of ways to select the first term $=n-4$
For $r=3$, the number of ways to select the first term $=n-6$
Now we see a pattern emerging, we also realize from this that $r<=\frac{n-1}{2}$ for an A.P. with 3 terms to exist in the given interval.
$\therefore$ The final answer $=n-2+n-4+n-6+...+5+3+1$
Now we use the formula to find the sum of an A.P. which is $S_n=\frac{n}{2}[a_1+a_n]$
$\therefore$ Answer $=\frac{n-1}{4}[1+n-2]=\frac{(n-1{)}^2}{4}$