If$3$ numbers are selected from the first $15$ natural numbers, then the probability that the numbers are in A.P. is 

Let $a<b<c$be in A.P. $\Rightarrow a+c=2b$, even number 
$\therefore$    a and c are both even or both odd. 
$\therefore$   Required probability $y=\frac{\left(7_{C_2}\right)+\left(8_{C_2}\right)}{\left({15}_{C_3}\right)}=\frac{21+28}{5\times 7\times 13}=\frac{7}{65}$