Find the value of p for which the numbers 2p-1, 3p+1, 11 are in AP. Hence, find the numbers.

Answer

To solve the problem, we need to find the value of \( p \) for which the numbers \( 2p - 1 \), \( 3p + 1 \), and \( 11 \) are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding the condition for AP**:
In an arithmetic progression, the difference between consecutive terms is constant. Thus, for three terms \( A \), \( B \), and \( C \) to be in AP, we have:
\[ B - A = C - B \]
Here, \( A = 2p - 1 \), \( B = 3p + 1 \), and \( C = 11 \).

2. **Setting up the equation**:
Using the condition for AP:
\[ (3p + 1) - (2p - 1) = 11 - (3p + 1) \]

3. **Simplifying the left side**:
\[ 3p + 1 - 2p + 1 = 11 - 3p - 1 \]
This simplifies to:
\[ p + 2 = 10 - 3p \]

4. **Rearranging the equation**:
Now, we can rearrange the equation to isolate \( p \):
\[ p + 3p + 2 = 10 \]
\[ 4p + 2 = 10 \]

5. **Solving for \( p \)**:
Subtract 2 from both sides:
\[ 4p = 8 \]
Now, divide by 4:
\[ p = 2 \]

6. **Finding the numbers**:
Now that we have \( p = 2 \), we can find the three numbers:
- First number:
\[ 2p - 1 = 2(2) - 1 = 4 - 1 = 3 \]
- Second number:
\[ 3p + 1 = 3(2) + 1 = 6 + 1 = 7 \]
- Third number (given):
\[ 11 \]

7. **Conclusion**:
The three numbers are \( 3 \), \( 7 \), and \( 11 \).

### Final Answer:
The value of \( p \) is \( 2 \) and the numbers are \( 3 \), \( 7 \), and \( 11 \).

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