Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :

Solution

Maths sir · 5 steps

Fix the first and last numbers; if they have the same parity, the middle term is decided for you.

1. Relabel the numbers. Shifting every number by the same amount keeps an A.P. an A.P., so take the 11 numbers as 1, 2, …, 11. Any 3 numbers chosen are unordered, so arrange them in increasing order. Then any A.P. among them has a positive common difference.
2. Total outcomes. Choosing 3 numbers out of 11 gives C(11, 3) = (11 × 10 × 9)/6 = 165 equally likely selections.
3. Favourable outcomes. Three numbers a < b < c are in A.P. when 2b = a + c. So a + c must be even, which means a and c are both odd or both even. Once a and c are picked, b = (a+c)/2 is fixed. There are 6 odd numbers (1, 3, 5, 7, 9, 11) and 5 even numbers (2, 4, 6, 8, 10).

   C(6, 2) + C(5, 2) = 15 + 10 = 25

4. Check by difference. Counting by common difference d gives the same total. The number of A.P.s with difference d is 11 - 2d, so d = 1, 2, 3, 4, 5 give 9 + 7 + 5 + 3 + 1 = 25.
5. So. The probability is 25/165 = 5/33, which is option (b).

The idea to keep. For three terms in A.P., choose the two end terms with the same parity and the middle term is forced. With an odd count of consecutive numbers, that is C(odds, 2) + C(evens, 2).

Answer from the NTA final answer key. Solution by PrepWiser.