Question

Given that $n$ is odd, the number of ways in which three numbers in $A$.$P$. can be selected from $1$, $2$, $3$, $4$, $.........n$ 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

Answer: (D)

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

Answer 2

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 even or both odd and this may be done in $^{\frac{n-1}{2}}C_{2}+^{\frac{n+1}{2}}C_{2}=\frac{\left(n-1\right)^{2}}{4}$ ways Note that, if $a$, $b$, $c$ are in $A$.$P$. $a + c = 2b$. Hence, if $a$, $b$, $c$ are integer the sum of extreme digits ($a$ and $c$) is even.