If k, `2k-1` and `2k+1` are three consecutive terms of an A.P., find the value of k.

Answer

To solve the problem, we need to find the value of \( k \) given that \( k \), \( 2k-1 \), and \( 2k+1 \) are three consecutive terms of an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the properties of A.P.**:
In an arithmetic progression, the difference between consecutive terms is constant. Therefore, we can set up the equation based on the first two terms and the last two terms.

2. **Setting up the equation**:
The first two terms are \( k \) and \( 2k-1 \). The common difference between these two terms can be expressed as:
\[ (2k - 1) - k \]
The last two terms are \( 2k-1 \) and \( 2k+1 \). The common difference between these two terms can be expressed as:
\[ (2k + 1) - (2k - 1) \]

3. **Calculating the differences**:
- For the first difference:
\[ (2k - 1) - k = 2k - 1 - k = k - 1 \]
- For the second difference:
\[ (2k + 1) - (2k - 1) = 2k + 1 - 2k + 1 = 2 \]

4. **Setting the differences equal**:
Since both differences must be equal in an A.P., we can set them equal to each other:
\[ k - 1 = 2 \]

5. **Solving for \( k \)**:
Now, we solve the equation:
\[ k - 1 = 2 \]
Adding 1 to both sides gives:
\[ k = 2 + 1 = 3 \]

6. **Conclusion**:
Thus, the value of \( k \) is \( 3 \).

### Final Answer:
\[ k = 3 \]