Final answer:
The chance that the numbers on the three drawn tickets are in an arithmetic progression (AP) is 3n / (4n²+2n-1). This probability is determined by analyzing the possible outcomes and calculating the probability of each favorable outcome.
Step-by-step explanation:
The probability that the numbers on the three drawn tickets are in an arithmetic progression (AP) can be determined by analyzing the possible outcomes and calculating the probability of each favorable outcome.
Let's consider a sample space of (2n+1) tickets consecutively numbered. To form an arithmetic progression, we need to select three tickets that are evenly spaced in terms of their numbers, such that the difference between consecutive numbers is the same.
The first ticket can be any of the (2n+1) tickets. The second ticket can be any of the remaining (2n) tickets. The third ticket can be any of the remaining (2n-1) tickets. Therefore, the total number of favorable outcomes is (2n+1)(2n)(2n-1).
The total number of possible outcomes is (2n+1)C3, which represents the combination of selecting three tickets from (2n+1) tickets.
Therefore, the probability of selecting three tickets in an AP is:
(2n+1)(2n)(2n-1) / [(2n+1)C3]
This simplifies to: 3n / (4n²+2n-1)