For what value of k will the consecutive terms `2k + 1, 3k+3`and `5k-1` form an AP?

Answer

To determine the value of \( k \) such that the terms \( 2k + 1 \), \( 3k + 3 \), and \( 5k - 1 \) form an arithmetic progression (AP), we can use the property of AP that states for three consecutive terms \( A \), \( B \), and \( C \) in AP, the following relationship holds:

\[ 2B = A + C \]

### Step-by-Step Solution:

1. **Identify the terms**:
- Let \( A = 2k + 1 \)
- Let \( B = 3k + 3 \)
- Let \( C = 5k - 1 \)

2. **Set up the equation using the AP property**:
According to the property of AP:
\[ 2B = A + C \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ 2(3k + 3) = (2k + 1) + (5k - 1) \]

3. **Simplify the left side**:
\[ 2(3k + 3) = 6k + 6 \]

4. **Simplify the right side**:
\[ (2k + 1) + (5k - 1) = 2k + 5k + 1 - 1 = 7k \]

5. **Set the two sides equal**:
\[ 6k + 6 = 7k \]

6. **Rearrange the equation**:
Subtract \( 6k \) from both sides:
\[ 6 = 7k - 6k \]
This simplifies to:
\[ 6 = k \]

7. **Conclusion**:
Thus, the value of \( k \) that makes the terms \( 2k + 1 \), \( 3k + 3 \), and \( 5k - 1 \) form an arithmetic progression is:
\[ k = 6 \]