Three Consecutive Even Numbers – UGC NET Question

Let the three consecutive even numbers be n − 2, n and n + 2, where n is the middle one. Their sum is (n − 2) + n + (n + 2) = 3n. According to the given condition, 3n = 66, so n = 66 ÷ 3 = 22. Hence, the middle even number is 22.

Option A:
Option A, 20, would give the three numbers 18, 20 and 22, whose sum is 60, not 66. Therefore, it does not satisfy the condition set in the problem.

Option B:
Option B fits perfectly: with 22 as the middle number, the three numbers become 20, 22 and 24, and their sum is 20 + 22 + 24 = 66, matching the given total.

Option C:
Option C, 24, would generate the sequence 22, 24 and 26, whose sum is 72, exceeding the required sum.

Option D:
Option D, 26, would produce 24, 26 and 28, summing to 78, which is also inconsistent with the problem statement.