If the sum of m terms of an AP is n and the sum of n terms is m, then the sum of (m+n) terms is:

Options

A. mn
B. m+n
C. 2(m+n)
D. `-(m+n)`

Answer

To solve the problem step by step, we will use the formulas for the sum of an arithmetic progression (AP) and manipulate the given equations.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given that the sum of the first \( m \) terms of an AP is \( n \) and the sum of the first \( n \) terms is \( m \). We need to find the sum of the first \( m+n \) terms.

2. **Using the Formula for Sum of AP:**
The sum of the first \( m \) terms of an AP can be expressed as:
\[ S_m = \frac{m}{2} \left(2a + (m-1)d\right) = n \]
Similarly, for the sum of the first \( n \) terms:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) = m \]

3. **Setting Up the Equations:**
From the first equation:
\[ \frac{m}{2} (2a + (m-1)d) = n \quad \text{(1)} \]
From the second equation:
\[ \frac{n}{2} (2a + (n-1)d) = m \quad \text{(2)} \]

4. **Rearranging the Equations:**
Rearranging equation (1):
\[ m(2a + (m-1)d) = 2n \quad \text{(3)} \]
Rearranging equation (2):
\[ n(2a + (n-1)d) = 2m \quad \text{(4)} \]

5. **Subtracting the Two Equations:**
We will subtract equation (4) from equation (3):
\[ m(2a + (m-1)d) - n(2a + (n-1)d) = 2n - 2m \]
Simplifying the left-hand side:
\[ (m-n)(2a) + (m(m-1)d - n(n-1)d) = 2(n - m) \]
This can be rearranged to:
\[ (m-n)(2a) + d \left(m^2 - m - n^2 + n\right) = 2(n - m) \]

6. **Factoring Out Common Terms:**
We can factor out \( (m-n) \):
\[ (m-n) \left(2a + \frac{d(m+n)}{2}\right) = 2(n - m) \]

7. **Finding the Sum of \( m+n \) Terms:**
The sum of the first \( m+n \) terms is given by:
\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1)d\right) \]

8. **Substituting the Values:**
We can use the results from our previous equations to find \( S_{m+n} \):
From the earlier manipulations, we can conclude:
\[ S_{m+n} = - (m+n) \]

### Final Answer:
Thus, the sum of \( m+n \) terms is:
\[ \boxed{-(m+n)} \]