Find the value of p, if x, 2x + p and 3x + 6 are in A.P.

Answer

To find the value of \( p \) such that \( x \), \( 2x + p \), and \( 3x + 6 \) are in Arithmetic Progression (A.P.), we can use the property of A.P. that states the difference between consecutive terms is constant.

### Step-by-Step Solution:

1. **Identify the terms in A.P.**:
- First term \( (t_1) = x \)
- Second term \( (t_2) = 2x + p \)
- Third term \( (t_3) = 3x + 6 \)

2. **Use the A.P. condition**:
For three terms to be in A.P., the following condition must hold:
\[ t_2 - t_1 = t_3 - t_2 \]
Substituting the values of \( t_1 \), \( t_2 \), and \( t_3 \):
\[ (2x + p) - x = (3x + 6) - (2x + p) \]

3. **Simplify both sides**:
- Left side:
\[ 2x + p - x = x + p \]
- Right side:
\[ 3x + 6 - 2x - p = x + 6 - p \]

4. **Set the two sides equal**:
Now we have:
\[ x + p = x + 6 - p \]

5. **Eliminate \( x \) from both sides**:
Subtract \( x \) from both sides:
\[ p = 6 - p \]

6. **Solve for \( p \)**:
Add \( p \) to both sides:
\[ p + p = 6 \]
\[ 2p = 6 \]
Divide both sides by 2:
\[ p = 3 \]

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

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