Question

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

• 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

Answer: (D)

$\frac{{(n-1)}^2}{4}$

There are in the set $1,2,3,...,n(n$ being odd $),\frac{n-1}{2}$ even numbers, $\frac{n+1}{2}$ odd numbers; and for an A.P.
the sum of the extremes is always even and hence the choice is either(both) $2$ even or (both) $2$ odd and this may be done in
${}^{\frac{n-1}{2}}{C}_2{+}^{\frac{n+1}{2}}{C}_2=\frac{{(n-1)}^2}{4}$ ways