If the sum of m terms of an A.P. is same as the sum of its n terms, then the sum of its (m+n) terms is

Options

A. mn
B. `-mn`
C. 1/mn
D. 0

Explanation

To solve the problem, we need to find the sum of (m+n) terms of an arithmetic progression (A.P.) given that the sum of m terms is equal to the sum of n terms.

### Step-by-Step Solution:

1. **Understanding the Sum of Terms in an A.P.**:
The sum of the first \( m \) terms of an A.P. can be expressed as:
\[ S_m = \frac{m}{2} \left(2a + (m-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

Similarly, the sum of the first \( n \) terms is:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

2. **Setting Up the Equation**:
According to the problem, we have:
\[ S_m = S_n \]
Thus, we can write:
\[ \frac{m}{2} \left(2a + (m-1)d\right) = \frac{n}{2} \left(2a + (n-1)d\right) \]

3. **Eliminating the Common Factor**:
We can multiply both sides by 2 to eliminate the fraction:
\[ m(2a + (m-1)d) = n(2a + (n-1)d) \]

4. **Expanding Both Sides**:
Expanding both sides gives:
\[ 2am + m(m-1)d = 2an + n(n-1)d \]

5. **Rearranging the Equation**:
Rearranging the terms leads to:
\[ 2am - 2an = n(n-1)d - m(m-1)d \]
Simplifying further:
\[ 2a(m-n) = d[n(n-1) - m(m-1)] \]

6. **Finding the Sum of (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 the Values**:
Since we know from our previous steps that:
\[ 2a + (m+n-1)d = 0 \]
(because \( 2a + (m+n-1)d \) would equal zero if \( 2a + (m-1)d = 2a + (n-1)d \)), we substitute this into the sum:
\[ S_{m+n} = \frac{m+n}{2} \cdot 0 = 0 \]

### Conclusion:
Thus, the sum of the first \( (m+n) \) terms of the A.P. is:
\[ \boxed{0} \]