Out of 21 tickets consecutively numbered, there are drawn at random. Find the probability that the numbers on them are in AP is a / b, then (14a - b) is ......... .

Any three ticket8 out of21 tickets can be chosen is  21C3 waysFor the favorable choice, if the chosen numbers are a, b and c, a < b < c then we should have a+c2=b.Obviously either both o and care even or both are odd and then 6 is fixed Hence, for the favorable choice, we have to choose two numbers from 1 to 21, which are either both even or both odd.This can be done in  11C2+10C2, ways. Hence, required probability = 11C2+10C2 21C3=10133⇒ab⇒14a−b=140−133=7