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If n is an odd positive integer, how many triples can be chosen from the set {1, 2, 3,..., n} so that the three selected numbers form an arithmetic progression?

1. (n−1)²/2
2. (n+1)²/4
3. (n+1)²/2
4. (n−1)²/4

Correct answer: (n−1)²/4

Solution

A 3-term AP is fixed by its first and last terms, which must have the same parity. For n = 2k+1 there are k+1 odds and k evens, giving C(k+1,2)+C(k,2) = k^2 pairs. With k = (n-1)/2 this equals (n-1)^2/4.

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