Suppose the sum of the first m teams of a arithmetic progression is n and the sum of its first n terms is m, where `mnen.` Then the sum of the first (m + n) terms of the arithmetic progression is

Options

A. 1- mn
B. mn- 5
C. `-(m = n)`
D. m + n

Explanation

To solve the problem, we need to find the sum of the first (m + n) terms of an arithmetic progression (AP) given that the sum of the first m terms is n and the sum of the first n terms is m, where m is not equal to n.

### Step-by-Step Solution:

1. **Understanding the Sum of an Arithmetic Progression**:
The sum of the first \( k \) terms of an arithmetic progression can be expressed as:
\[ S_k = \frac{k}{2} \left(2a + (k - 1)d\right) \]
where \( a \) is the first term, \( d \) is the common difference, and \( k \) is the number of terms.

2. **Setting Up the Equations**:
From the problem, we have:
- The sum of the first \( m \) terms is \( n \):
\[ S_m = \frac{m}{2} \left(2a + (m - 1)d\right) = n \quad \text{(Equation 1)} \]
- The sum of the first \( n \) terms is \( m \):
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) = m \quad \text{(Equation 2)} \]

3. **Rearranging the Equations**:
From Equation 1:
\[ m(2a + (m - 1)d) = 2n \quad \text{(Multiply both sides by 2)} \]
From Equation 2:
\[ n(2a + (n - 1)d) = 2m \quad \text{(Multiply both sides by 2)} \]

4. **Subtracting the Two Equations**:
Now, we subtract Equation 1 from Equation 2:
\[ n(2a + (n - 1)d) - m(2a + (m - 1)d) = 2m - 2n \]
Simplifying gives:
\[ n(2a + (n - 1)d) - m(2a + (m - 1)d) = 2(m - n) \]

5. **Factoring Out Common Terms**:
Rearranging the left side:
\[ (n - m) \cdot 2a + (n(n - 1) - m(m - 1))d = 2(m - n) \]
This can be simplified to:
\[ (n - m)(2a + d) = 2(m - n) \]

6. **Finding the Sum of the First (m + n) Terms**:
Now, we need to find the sum of the first \( m + n \) terms:
\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1)d\right) \]

7. **Substituting Values**:
We can express \( 2a + (m+n-1)d \) in terms of \( m \) and \( n \):
\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1)d\right) \]
Using the previous results, we can substitute \( 2a + (m+n-1)d \) based on the derived equations.

8. **Final Result**:
After substituting and simplifying, we find that:
\[ S_{m+n} = - (m + n) \]

### Conclusion:
Thus, the sum of the first \( m + n \) terms of the arithmetic progression is:
\[ \boxed{-(m + n)} \]