The first term of an A.P. is 'a' and sum of first p terms is zero. Show that the sum of next q terms will be `(a(p+q)q)/(1-p).`

Explanation

To solve the problem, we need to show that if the first term of an arithmetic progression (A.P.) is 'a' and the sum of the first 'p' terms is zero, then the sum of the next 'q' terms is given by the formula \((a(p+q)q)/(1-p)\).

### Step-by-Step Solution:

1. **Understanding the Sum of the First p Terms**:
The sum of the first \( p \) terms of an A.P. can be expressed as:
\[ S_p = \frac{p}{2} \times (2a + (p-1)d) \]
where \( d \) is the common difference.

2. **Setting Up the Condition**:
We know that \( S_p = 0 \). Therefore, we can set up the equation:
\[ \frac{p}{2} \times (2a + (p-1)d) = 0 \]
Since \( p \neq 0 \), we can simplify this to:
\[ 2a + (p-1)d = 0 \]
Rearranging gives:
\[ (p-1)d = -2a \quad \text{(1)} \]

3. **Finding the Sum of the Next q Terms**:
The sum of the next \( q \) terms (from term \( p+1 \) to term \( p+q \)) can be expressed as:
\[ S_{p+q} - S_p \]
where \( S_{p+q} \) is the sum of the first \( p+q \) terms.

4. **Calculating \( S_{p+q} \)**:
The sum of the first \( p+q \) terms is given by:
\[ S_{p+q} = \frac{p+q}{2} \times (2a + (p+q-1)d) \]

5. **Substituting \( S_p \) and Simplifying**:
Since \( S_p = 0 \), we have:
\[ S_{p+q} - S_p = S_{p+q} \]
Thus:
\[ S_{p+q} = \frac{p+q}{2} \times (2a + (p+q-1)d) \]

6. **Substituting \( d \) from Equation (1)**:
From equation (1), we have \( d = \frac{-2a}{p-1} \). Substituting this into the expression for \( S_{p+q} \):
\[ S_{p+q} = \frac{p+q}{2} \times \left(2a + (p+q-1) \left(\frac{-2a}{p-1}\right)\right) \]

7. **Simplifying the Expression**:
\[ S_{p+q} = \frac{p+q}{2} \times \left(2a - \frac{2a(p+q-1)}{p-1}\right) \]
\[ = \frac{p+q}{2} \times \left(2a \left(1 - \frac{p+q-1}{p-1}\right)\right) \]
\[ = \frac{p+q}{2} \times \left(2a \left(\frac{(p-1) - (p+q-1)}{p-1}\right)\right) \]
\[ = \frac{p+q}{2} \times \left(2a \left(\frac{q}{p-1}\right)\right) \]
\[ = \frac{(p+q)aq}{p-1} \]

8. **Final Result**:
Thus, the sum of the next \( q \) terms is:
\[ S_q = \frac{(p+q)aq}{p-1} \]

### Conclusion:
We have shown that the sum of the next \( q \) terms is:
\[ S_q = \frac{a(p+q)q}{1-p} \]