Statement-1: `n(n+1)(n+2)` is always divisible by 6 for all `n in N`.
Statement-2: The product of any two consecutive natural number is divisible by 2.

Answer

To determine whether the statements are true, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement 1:** \( n(n+1)(n+2) \) is always divisible by 6 for all \( n \in \mathbb{N} \).

1. **Understanding divisibility by 6:**
- A number is divisible by 6 if it is divisible by both 2 and 3.

2. **Check divisibility by 2:**
- The product \( n(n+1)(n+2) \) consists of three consecutive integers.
- Among any three consecutive integers, at least one of them must be even. Therefore, \( n(n+1)(n+2) \) is always divisible by 2.

3. **Check divisibility by 3:**
- Among any three consecutive integers, at least one of them must be divisible by 3. This is because the integers can be expressed in the forms of \( 3k \), \( 3k+1 \), and \( 3k+2 \) for some integer \( k \).
- Hence, \( n(n+1)(n+2) \) is also divisible by 3.

4. **Conclusion for Statement 1:**
- Since \( n(n+1)(n+2) \) is divisible by both 2 and 3, it is therefore divisible by 6.
- Thus, Statement 1 is **true**.

### Step 2: Analyze Statement 2
**Statement 2:** The product of any two consecutive natural numbers is divisible by 2.

1. **Understanding consecutive natural numbers:**
- Two consecutive natural numbers can be represented as \( n \) and \( n+1 \).

2. **Check divisibility by 2:**
- Among any two consecutive integers, one must be even (since even and odd alternate).
- Therefore, the product \( n(n+1) \) will always include at least one even number.

3. **Conclusion for Statement 2:**
- Since the product of an even number with any integer is always even, \( n(n+1) \) is divisible by 2.
- Thus, Statement 2 is also **true**.

### Final Conclusion:
Both statements are true:
- **Statement 1:** True
- **Statement 2:** True

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