The sum of three consecutive odd numbers is always divisible by I. 2 II. 3 III. 5 IV. 6

## Concept & Logic The sum of any arithmetic progression with an odd number of terms is equal to the middle term multiplied by the number of terms. ## Step-by-Step Solution * Let the three consecutive odd numbers be $(2k - 1)$, $(2k + 1)$, and $(2k + 3)$ for some integer $k$. * Sum these expressions: $$ (2k - 1) + (2k + 1) + (2k + 3) = 6k + 3 $$ * Factor out the common term: $$ 3(2k + 1) $$ * Because 3 is a multiplier, the sum is always a multiple of 3. * Since $(2k + 1)$ is always odd, the sum is an odd multiple of 3. Therefore, it will never be divisible by an even number like 2 or 6. ## Exam Strategy & Shortcut Substitute the smallest consecutive odd numbers to quickly eliminate options. Try 1, 3, and 5. Their sum is $1 + 3 + 5 = 9$. 9 is only divisible by 3 (Statement II). It is not divisible by 2, 5, or 6. ## Common Pitfall Students often mistakenly assume that because the sum formula yields $6k + 3$, they should look for divisibility by 6. However, adding odd numbers always results in an odd sum, making divisibility by any even number impossible. ## Final Answer **Therefore, the correct answer is Only II.**