If in an A.P. the ratio of the sum of m terms and n terms is `m^(2) : n^(2)` then if a is first term and d the common difference, then

Options

A. a = 2d
B. a = d
C. d = 2a
D. none of these

Answer

To solve the problem, we need to establish the relationship between the first term \( a \), the common difference \( d \), and the number of terms \( m \) and \( n \) in an arithmetic progression (A.P.) given that the ratio of the sums of the first \( m \) terms and the first \( n \) terms is \( \frac{m^2}{n^2} \).

### Step-by-Step Solution:

1. **Write the formula for the sum of the first \( m \) terms of an A.P.**:
The sum \( S_m \) of the first \( m \) terms is given by:
\[ S_m = \frac{m}{2} \left(2a + (m - 1)d\right) \]
Similarly, the sum \( S_n \) of the first \( n \) terms is:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]

2. **Set up the equation based on the given ratio**:
According to the problem, we have:
\[ \frac{S_m}{S_n} = \frac{m^2}{n^2} \]
Substituting the formulas for \( S_m \) and \( S_n \):
\[ \frac{\frac{m}{2} \left(2a + (m - 1)d\right)}{\frac{n}{2} \left(2a + (n - 1)d\right)} = \frac{m^2}{n^2} \]

3. **Simplify the equation**:
The \( \frac{1}{2} \) cancels out:
\[ \frac{m(2a + (m - 1)d)}{n(2a + (n - 1)d)} = \frac{m^2}{n^2} \]

4. **Cross-multiply to eliminate the fraction**:
\[ m(2a + (m - 1)d) \cdot n^2 = n(2a + (n - 1)d) \cdot m^2 \]

5. **Expand both sides**:
\[ mn^2(2a + (m - 1)d) = nm^2(2a + (n - 1)d) \]

6. **Rearranging the equation**:
Expanding both sides gives:
\[ 2amn^2 + m(n^2)(m - 1)d = 2amn^2 + n(m^2)(n - 1)d \]
Canceling \( 2amn^2 \) from both sides:
\[ m(n^2)(m - 1)d = n(m^2)(n - 1)d \]

7. **Factor out \( d \)** (assuming \( d \neq 0 \)):
\[ m(n^2)(m - 1) = n(m^2)(n - 1) \]

8. **Rearranging gives us**:
\[ mn^2(m - 1) = nm^2(n - 1) \]

9. **Dividing both sides by \( mn \)** (assuming \( m, n \neq 0 \)):
\[ n(m - 1) = m(n - 1) \]

10. **Expanding and simplifying**:
\[ nm - n = mn - m \]
This simplifies to:
\[ m = n \]

11. **Final result**:
From the earlier steps, we derived that \( 2a = d \).

### Conclusion:
Thus, the final result is:
\[ 2a = d \]