An arithmetic progression P consists of n terms. From the progression three different progressions `P_1P_2 and P_3` are created such that `P_1` is obtained by the 1st, 4th, 7th .... terms of P, `P_2` has the 2nd, 5th, 8th, ..... terms of P and `P_3` has the 3rd, 6th, 9th, ..... terms of P. It is found that of P1, P2 and P3 two progressions have the property that their average is itself a term of the original Progression P. Which of the following can be a possible value of n?

Options

A. 20
B. 26
C. 36
D. both a and b

Answer

To solve the problem, we need to analyze the given arithmetic progression \( P \) and the derived progressions \( P_1, P_2, \) and \( P_3 \).

### Step-by-Step Solution:

1. **Understanding the Progressions**:
- The original arithmetic progression \( P \) consists of \( n \) terms.
- The terms of \( P_1 \) are the 1st, 4th, 7th, ..., which can be expressed as \( P_1 = a, a + 3d, a + 6d, \ldots \) where \( a \) is the first term and \( d \) is the common difference.
- The terms of \( P_2 \) are the 2nd, 5th, 8th, ..., which can be expressed as \( P_2 = a + d, a + 4d, a + 7d, \ldots \).
- The terms of \( P_3 \) are the 3rd, 6th, 9th, ..., which can be expressed as \( P_3 = a + 2d, a + 5d, a + 8d, \ldots \).

2. **Counting the Terms**:
- The number of terms in \( P_1 \) can be calculated as \( \lceil \frac{n}{3} \rceil \).
- The number of terms in \( P_2 \) is also \( \lceil \frac{n-1}{3} \rceil \).
- The number of terms in \( P_3 \) is \( \lceil \frac{n-2}{3} \rceil \).

3. **Finding Averages**:
- For \( P_1 \), if it has \( k_1 \) terms, the average will be \( \frac{(k_1)(a + (k_1 - 1) \cdot 3d)}{k_1} = a + (k_1 - 1) \cdot \frac{3d}{2} \).
- For \( P_2 \), if it has \( k_2 \) terms, the average will be \( a + d + (k_2 - 1) \cdot \frac{3d}{2} \).
- For \( P_3 \), if it has \( k_3 \) terms, the average will be \( a + 2d + (k_3 - 1) \cdot \frac{3d}{2} \).

4. **Condition for Averages**:
- The problem states that two of the averages must be terms of the original progression \( P \). This means that the average must equal \( a + md \) for some integer \( m \).
- We need to check which values of \( n \) allow for this condition.

5. **Testing the Options**:
- **Option A: \( n = 20 \)**:
- \( P_1 \) has 7 terms, \( P_2 \) has 7 terms, \( P_3 \) has 6 terms.
- The averages of \( P_1 \) and \( P_2 \) can be calculated and checked if they belong to \( P \). It turns out they do.
- **Option B: \( n = 26 \)**:
- \( P_1 \) has 9 terms, \( P_2 \) has 9 terms, \( P_3 \) has 8 terms.
- The averages of \( P_1 \) and \( P_2 \) can also be checked, and they satisfy the condition.
- **Option C: \( n = 36 \)**:
- \( P_1 \), \( P_2 \), and \( P_3 \) all have 12 terms.
- The averages will not yield a term in \( P \) as they are all even, and thus do not satisfy the condition.
- **Conclusion**: The possible values of \( n \) that satisfy the condition are \( 20 \) and \( 26 \).

### Final Answer:
The possible values of \( n \) are **20 and 26**.