If n is an odd number, how many ways can three terms in an arithmetic progression be chosen from the sequence 1, 2, 3,..., n?

Question

Accepted Answer

(n - 1)² / 4. A 3-term AP (a, a+d, a+2d) from 1..n with common difference d>=1 needs a+2d<=n, giving n-2d choices, summed over d=1..(n-1)/2. The total is sum(n-2d) = ((n-1)/2)^2 = (n-1)^2/4. Check n=5: d=1 gives 3, d=2 gives 1, total 4 = (5-1)^2/4 = 4. So the answer is (n-1)^2/4 (index 3), not (n+1)^2/2; stored index 1 is wrong.

If n is an odd number, how many ways can three terms in an arithmetic progression be chosen from the sequence 1, 2, 3,..., n?

Solution

Related JEE Advanced Maths questions