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 A.P.**:
The sum of the first \( k \) terms of an A.P. 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 Given Condition**:
According to the problem, the sum of the first \( m \) terms is equal to the sum of the first \( n \) terms:
\[ S_m = S_n \]
Therefore, 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:
\[ 2am + md(m-1) = 2an + nd(n-1) \]

5. **Rearranging the Equation**:
Rearranging the equation, we get:
\[ 2am - 2an + md(m-1) - nd(n-1) = 0 \]
This can be simplified to:
\[ 2a(m - n) + d \left[m(m-1) - n(n-1)\right] = 0 \]

6. **Factoring the Equation**:
We can factor out \( (m-n) \):
\[ (m-n) \left(2a + d(m+n-1)\right) = 0 \]

7. **Analyzing the Factors**:
Since \( m \) and \( n \) are not equal (as stated in the problem), we have:
\[ 2a + d(m+n-1) = 0 \]
This implies:
\[ 2a = -d(m+n-1) \]

8. **Finding the Sum of \( (m+n) \) Terms**:
Now, we need to find the sum of the first \( (m+n) \) terms:
\[ S_{m+n} = \frac{m+n}{2} \left(2a + (m+n-1)d\right) \]
Substituting \( 2a \) from the previous step:
\[ S_{m+n} = \frac{m+n}{2} \left(-d(m+n-1) + (m+n-1)d\right) \]
This simplifies to:
\[ S_{m+n} = \frac{m+n}{2} \cdot 0 = 0 \]

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