Final answer:
The total number of ways three distinct numbers in AP can be selected from {1, 2, 3, …, 24} is determined by counting possible sequences based on choices for the first term and the common difference. The answer is 66 ways.
Step-by-step explanation:
The question asks about the total number of ways in which three distinct numbers in Arithmetic Progression (AP) can be selected from the set {1, 2, 3, …, 24}. To form an arithmetic progression, we choose the first term (a), the common difference (d), and the third term will be (a + 2d). The third term must not exceed 24. For each choice of a and d, there is exactly one AP formed. To find the number of APs, we look at the largest possible value for the first term, which would be 22, because the second term would then be 23 and the third term would be 24. For a=1, we could choose d from 1 up to 11 (which would yield the sequence 1, 12, 23). As we increase the first term, the number of possible differences decreases. The total number of APs is the sum of the first 11 positive integers which is ½(11)(12) which equals 66. Therefore, the answer is (a) 66.