Three numbers are selected from the set {1, 2, 3, …., 23, 24}, with out replacement. The probability that the formed numbers from an A.P. is equal to

$\frac{3}{46}$

Let the selected numbers be n₁, n₂ and n₃.

We must have

2n₂ = n₁ + n₃ 

Thus n₁ + n₂ must be even. That means n₁ and n₃ both must have same nature.

Thus, required probability = $\frac{2 .^{12} C_2}{{ }^{24} C_3}=\frac{3}{46}$