The pth term of an A.P. is q and the qth term is p, show that the mth term is p + q -m.

Answer

To solve the problem, we need to show that the mth term of an arithmetic progression (A.P.) is given by the expression \( p + q - m \) given that the pth term is \( q \) and the qth term is \( p \).

### Step-by-Step Solution:

1. **Understanding the nth term of an A.P.**:
The nth term of an A.P. can be expressed as:
\[ a_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Setting up the equations**:
From the problem statement, we have:
- The pth term is \( q \):
\[ a + (p - 1)d = q \quad \text{(1)} \]
- The qth term is \( p \):
\[ a + (q - 1)d = p \quad \text{(2)} \]

3. **Rearranging the equations**:
From equation (1):
\[ a = q - (p - 1)d \quad \text{(3)} \]
From equation (2):
\[ a = p - (q - 1)d \quad \text{(4)} \]

4. **Equating the two expressions for \( a \)**:
Set equations (3) and (4) equal to each other:
\[ q - (p - 1)d = p - (q - 1)d \]

5. **Simplifying the equation**:
Rearranging gives:
\[ q - p + d(p - 1) = d(q - 1) \]
\[ q - p = d(q - p) \]
If \( q \neq p \), we can divide both sides by \( q - p \):
\[ 1 = d \]
Thus, we find \( d = 1 \).

6. **Substituting \( d \) back to find \( a \)**:
Substitute \( d = 1 \) into either equation (3) or (4). Using equation (3):
\[ a = q - (p - 1) \cdot 1 = q - p + 1 \]

7. **Finding the mth term**:
Now, we can find the mth term \( a_m \):
\[ a_m = a + (m - 1)d \]
Substitute \( a \) and \( d \):
\[ a_m = (q - p + 1) + (m - 1) \cdot 1 \]
Simplifying this:
\[ a_m = q - p + 1 + m - 1 = q - p + m \]

8. **Rearranging to match the required form**:
Rearranging gives:
\[ a_m = p + q - m \]

Thus, we have shown that the mth term of the A.P. is indeed \( p + q - m \).

### Final Result:
\[ \text{The mth term is } p + q - m. \]