If in an AP, `S_(n)= qn^(2)` and ` S_(m) =qm^(2)` , where `S_(r)` denotes the of r terms of the AP , then `S_(q) ` equals to

Options

A. `(q^(3))/(2)`
B. mnq
C. `q^(3)`
D. `(m+n)q^(2)`

Explanation

To solve the problem, we need to find \( S_q \) given that \( S_n = qn^2 \) and \( S_m = qm^2 \), where \( S_r \) denotes the sum of the first \( r \) terms of an arithmetic progression (AP).

### Step-by-step Solution:

1. **Understanding the Sum of Terms**:
The sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Equating Given Sums**:
From the problem, we have:
\[ S_n = qn^2 \]
and
\[ S_m = qm^2 \]

3. **Finding \( S_1 \), \( S_2 \), \( S_3 \), and \( S_4 \)**:
- For \( n = 1 \):
\[ S_1 = q(1^2) = q \]
- For \( n = 2 \):
\[ S_2 = q(2^2) = 4q \]
- For \( n = 3 \):
\[ S_3 = q(3^2) = 9q \]
- For \( n = 4 \):
\[ S_4 = q(4^2) = 16q \]

4. **Finding the Terms of the AP**:
The \( n \)-th term \( T_n \) can be found using:
\[ T_n = S_n - S_{n-1} \]
- For \( n = 2 \):
\[ T_2 = S_2 - S_1 = 4q - q = 3q \]
- For \( n = 3 \):
\[ T_3 = S_3 - S_2 = 9q - 4q = 5q \]
- For \( n = 4 \):
\[ T_4 = S_4 - S_3 = 16q - 9q = 7q \]

5. **Identifying the Sequence**:
The terms we have found are:
- \( T_1 = q \)
- \( T_2 = 3q \)
- \( T_3 = 5q \)
- \( T_4 = 7q \)

This indicates that the first term \( a = q \) and the common difference \( d = 2q \).

6. **Finding \( S_q \)**:
Now, we need to find \( S_q \):
\[ S_q = \frac{q}{2} \left(2a + (q-1)d\right) \]
Substituting \( a \) and \( d \):
\[ S_q = \frac{q}{2} \left(2q + (q-1)(2q)\right) \]
Simplifying:
\[ S_q = \frac{q}{2} \left(2q + 2q^2 - 2q\right) = \frac{q}{2} \cdot 2q^2 = q^3 \]

### Final Answer:
Thus, \( S_q = q^3 \).