Let `m in N` and suppose three numbers are chosen at random from the numbers 1,2,3,..,m.
Statement-1 : If m=2n fro some `n in N`, then the chosen numbers are in A. P. with probability `(3)/(2(2n-1))`.
Statement-2 : If m=2n+1, then the chosen numbers are in A.P. with probability `(3n)/(4n^(2)-1)`

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze the two statements regarding the probability of choosing three numbers from the set {1, 2, 3, ..., m} such that they are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

**Step 1: Analyze Statement 1**
- Given: \( m = 2n \) for some \( n \in \mathbb{N} \).
- We need to find the probability that three randomly chosen numbers from the set {1, 2, ..., m} are in A.P.
- The total ways to choose 3 numbers from \( m \) numbers is given by \( \binom{m}{3} \).

**Step 2: Calculate Total Combinations**
- For \( m = 2n \), the total combinations are:
\[ \binom{2n}{3} = \frac{2n(2n-1)(2n-2)}{6} = \frac{n(2n-1)(2n-2)}{3} \]

**Step 3: Identify Favorable Outcomes**
- The three numbers \( a, b, c \) are in A.P. if \( 2b = a + c \).
- The possible sets of three numbers that can be in A.P. can be derived from the pairs of numbers:
- For \( m = 2n \), the valid A.P. triplets are:
- \( (1, 2, 3) \)
- \( (2, 3, 4) \)
- ...
- \( (2n-2, 2n-1, 2n) \)
- The total number of A.P. triplets is \( n - 1 \) (as there are \( n-1 \) valid middle elements).

**Step 4: Calculate Probability for Statement 1**
- The probability \( P \) that three chosen numbers are in A.P. is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{n-1}{\binom{2n}{3}} \]
- After substituting and simplifying, we find:
\[ P = \frac{3}{2(2n-1)} \]

**Step 5: Analyze Statement 2**
- Given: \( m = 2n + 1 \).
- The total ways to choose 3 numbers from \( m \) numbers is:
\[ \binom{2n+1}{3} = \frac{(2n+1)(2n)(2n-1)}{6} \]

**Step 6: Identify Favorable Outcomes for Statement 2**
- Similar to the previous case, we can find the valid A.P. triplets:
- The valid A.P. triplets can be derived from the pairs of numbers:
- For \( m = 2n + 1 \), the total number of A.P. triplets is \( n \).

**Step 7: Calculate Probability for Statement 2**
- The probability \( P \) that three chosen numbers are in A.P. is:
\[ P = \frac{n}{\binom{2n+1}{3}} \]
- After substituting and simplifying, we find:
\[ P = \frac{3n}{4n^2 - 1} \]

### Conclusion:
- **Statement 1** is true with probability \( \frac{3}{2(2n-1)} \).
- **Statement 2** is true with probability \( \frac{3n}{4n^2 - 1} \).
- Both statements are true, but Statement 2 does not explain Statement 1.