If 3k - 2, 4k - 6 and k + 2 are three consecutive terms of A.P., then find the value of k.

Answer

To solve the problem, we need to find the value of \( k \) such that the terms \( 3k - 2 \), \( 4k - 6 \), and \( k + 2 \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Terms**:
The three terms are:
- First term: \( a_1 = 3k - 2 \)
- Second term: \( a_2 = 4k - 6 \)
- Third term: \( a_3 = k + 2 \)

2. **Use the Property of A.P.**:
For three terms to be in A.P., the difference between the first and second terms must equal the difference between the second and third terms:
\[ a_2 - a_1 = a_3 - a_2 \]

3. **Set Up the Equation**:
Substitute the terms into the equation:
\[ (4k - 6) - (3k - 2) = (k + 2) - (4k - 6) \]

4. **Simplify the Left Side**:
\[ 4k - 6 - 3k + 2 = k - 4 \]
This simplifies to:
\[ k - 4 \]

5. **Simplify the Right Side**:
\[ k + 2 - 4k + 6 = -3k + 8 \]

6. **Set the Two Sides Equal**:
Now we have:
\[ k - 4 = -3k + 8 \]

7. **Solve for \( k \)**:
Add \( 3k \) to both sides:
\[ k + 3k - 4 = 8 \]
This simplifies to:
\[ 4k - 4 = 8 \]
Now add \( 4 \) to both sides:
\[ 4k = 12 \]
Finally, divide by \( 4 \):
\[ k = 3 \]

### Final Answer:
The value of \( k \) is \( 3 \).