The total number of ways in which three distinct numbers in Arithmetic Progression can be selected from the set  1,2,3,.......,24 is equal to

Let the numbers selected be x1,x2,x3 We must have 2x2=x1+x3⇒x1+x3= even  Therefore both x1,x3 are odd (or) both are even.  If x1 and x3 both are even, we can select them in 12C2 ways.  Similarly, if both x1 and x3 are odd,  we can again select them in 12C2 ways.  Thus, the total number of ways is 2×12C2=132