In an arithmetical progression, the sum of p terms is m and the sum of q terms is also m. Find the sum of (p `+` q) terms.

Answer

To solve the problem, we need to find the sum of \( (p + q) \) terms in an arithmetic progression (AP) given that the sum of \( p \) terms is \( m \) and the sum of \( q \) terms is also \( m \).

### Step-by-Step Solution:

1. **Understanding the Sum of Terms in AP**:
The sum of the first \( n \) terms of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

2. **Sum of \( p \) Terms**:
Given that the sum of the first \( p \) terms is \( m \):
\[ S_p = \frac{p}{2} \times (2a + (p - 1)d) = m \]
Rearranging gives:
\[ 2m = p(2a + (p - 1)d) \quad \text{(Equation 1)} \]

3. **Sum of \( q \) Terms**:
Similarly, for the sum of the first \( q \) terms, we have:
\[ S_q = \frac{q}{2} \times (2a + (q - 1)d) = m \]
Rearranging gives:
\[ 2m = q(2a + (q - 1)d) \quad \text{(Equation 2)} \]

4. **Setting Up the Equations**:
From Equations 1 and 2, we can express both equations as:
\[ 2a + (p - 1)d = \frac{2m}{p} \]
\[ 2a + (q - 1)d = \frac{2m}{q} \]

5. **Subtracting the Two Equations**:
Subtract Equation 2 from Equation 1:
\[ (p - 1)d - (q - 1)d = \frac{2m}{p} - \frac{2m}{q} \]
This simplifies to:
\[ (p - q)d = 2m \left( \frac{1}{p} - \frac{1}{q} \right) \]
Simplifying the right-hand side:
\[ (p - q)d = 2m \left( \frac{q - p}{pq} \right) \]
Thus:
\[ d = -\frac{2m}{pq} \]

6. **Finding the Value of \( a \)**:
Substitute \( d \) back into either Equation 1 or Equation 2 to find \( a \). Using Equation 1:
\[ 2a + (p - 1)\left(-\frac{2m}{pq}\right) = \frac{2m}{p} \]
Rearranging gives:
\[ 2a = \frac{2m}{p} + \frac{2m(p - 1)}{pq} \]
Simplifying further:
\[ 2a = \frac{2m}{p} + \frac{2m(p - 1)}{pq} = \frac{2m(q + p - 1)}{pq} \]
Thus:
\[ a = \frac{m(q + p - 1)}{pq} \]

7. **Finding the Sum of \( (p + q) \) Terms**:
Now we can find the sum of \( (p + q) \) terms:
\[ S_{p+q} = \frac{p + q}{2} \times \left( 2a + (p + q - 1)d \right) \]
Substitute \( a \) and \( d \):
\[ S_{p+q} = \frac{p + q}{2} \left( 2 \times \frac{m(q + p - 1)}{pq} + (p + q - 1)\left(-\frac{2m}{pq}\right) \right) \]
Simplifying the expression inside the brackets:
\[ = \frac{p + q}{2} \left( \frac{2m(q + p - 1) - 2m(p + q - 1)}{pq} \right) \]
The terms cancel out:
\[ = \frac{p + q}{2} \times 0 = 0 \]

### Final Result:
Thus, the sum of \( (p + q) \) terms is:
\[ \boxed{0} \]