Question

The number of ways in which $3$ numbers in $A.P.$ can be selected from $1,2,3,...,n$ is

• A. $\frac{1}{4}(n-1{)}^2$ if $n$ is odd
• B. $\frac{1}{4}n(n-2)$ if $n$ is even
• C. $\frac{1}{2}n(n-2)$ if $n$ is even
• D. $\frac{1}{2}(n-1{)}^2$ if $n$ is odd

Answer 1

Answer: (A), (B)

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

The given numbers are $1,2,3,....n$

Let the numbers chosen be $a,b,c$

$a,b,c$ should be in A.P.
$\Rightarrow b=\frac{a+c}{2}\Rightarrow 2b=a+c$
$\therefore a+c$ is even
So, it should be either both even or both odd
Case 1:
$n$ is even
$\Rightarrow$ Number of odd numbers $=m=$ Number of even numbers
$=\frac{n}{2}$
So, number of selections $=2\cdot {}^m{C}_2=m(m-1)=\frac{n(n-2)}{4}$
Case 2:
$n$ is odd$(n=2m+1)$

$\Rightarrow$ Number of odd numbers $=m+1$ and number of even number $=m$

$={}^{m+1}{C}_2+{}^m{C}_2$

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