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Number of ways in which three numbers in...

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

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.

Text Solution

Verified by Experts

The correct Answer is:
B, C
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]`
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