If `k+3, 5k-1` and `3k+2` are in A.P., find the value of k.

Answer

To find the value of \( k \) such that the terms \( k + 3 \), \( 5k - 1 \), and \( 3k + 2 \) are in Arithmetic Progression (A.P.), we can use the property of A.P. that states:

For three terms \( a_1, a_2, a_3 \) to be in A.P., the following condition must hold:
\[ a_2 - a_1 = a_3 - a_2 \]

### Step-by-Step Solution:

1. **Identify the terms**:
- Let \( a_1 = k + 3 \)
- Let \( a_2 = 5k - 1 \)
- Let \( a_3 = 3k + 2 \)

2. **Set up the equation using the A.P. property**:
\[ a_2 - a_1 = a_3 - a_2 \]
Substituting the values:
\[ (5k - 1) - (k + 3) = (3k + 2) - (5k - 1) \]

3. **Simplify the left-hand side**:
\[ 5k - 1 - k - 3 = 4k - 4 \]

4. **Simplify the right-hand side**:
\[ 3k + 2 - 5k + 1 = -2k + 3 \]

5. **Set the two sides equal**:
\[ 4k - 4 = -2k + 3 \]

6. **Combine like terms**:
- Add \( 2k \) to both sides:
\[ 4k + 2k - 4 = 3 \]
This simplifies to:
\[ 6k - 4 = 3 \]

7. **Add 4 to both sides**:
\[ 6k = 3 + 4 \]
This simplifies to:
\[ 6k = 7 \]

8. **Divide by 6**:
\[ k = \frac{7}{6} \]

### Final Answer:
The value of \( k \) is \( \frac{7}{6} \).