Number of ways in which three numbers in AP can be selected from 1,2,3, . .,n is

Options

A. `((n-1)/(2))^(2)`, if n is even
B. `(n(n-2))/(4)`, if n is even
C. `((n-1)^(2))/(4)`, if n is odd
D. `(n(n+1))/(2)`, if n is odd.

Answer

B, C

Explanation

If a,b,c and in AP, then a+c=2b
a and b both are odd or both are even.
Case I if n is even
Let n=2m in which m are evenn annd m are odd numbers.
`therefore`Number of ways`=.^(m)C_(2)+.^(m)C_(2)=2*.^(m)C_(2)=2*(m(m-1))/(2)`
`=(n)/(2)((n)/(2)-1)=(n(n-2))/(4)" "[becausen=2m]`
Case II If n is odd
Let n=2m+1 in which m are even and m+1 are odd numbers.
`therefore`Number of ways`=.^(m)C_(2)+.^(m+1)C_(2)`
`(m(m-1))/(2)+((m+1)m)/(2)=m^(2)=((n-1)^(2))/(4)" "[becausen=2m+1]`