Consider an AP with a as the first term and d is the common difference such that `S_(n)` denotes the sum to n terms and `a_(n)` denotes the nth term of the AP. Given that for some m,`n inN,(S_(m))/(S_(n))=(m^(2))/(n^(2))(nen)`.
Statement 1 `d=2a` because
Statement 2 `(a_(m))/(a_(n))=(2m+1)/(2n+1)`.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a corrct explanation for Statement 1.
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
C. Statement 1 is true, Statement 2 is false.
D. Statement 1 is false, Statement 2 is true.

Explanation

To solve the problem, we need to analyze the given information about the arithmetic progression (AP) and the statements provided.

### Step-by-Step Solution:

1. **Understanding the Given Ratio**:
We are given that:
\[ \frac{S_m}{S_n} = \frac{m^2}{n^2} \]
where \( S_n \) is the sum of the first \( n \) terms of the AP.

2. **Formula for the Sum of an AP**:
The sum of the first \( n \) terms of an AP can be expressed as:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

3. **Expressing \( S_m \) and \( S_n \)**:
Using the formula for the sum of an AP, we can write:
\[ S_m = \frac{m}{2} \left(2a + (m-1)d\right) \]
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

4. **Setting Up the Ratio**:
Plugging these into the ratio gives:
\[ \frac{S_m}{S_n} = \frac{\frac{m}{2} \left(2a + (m-1)d\right)}{\frac{n}{2} \left(2a + (n-1)d\right)} = \frac{m(2a + (m-1)d)}{n(2a + (n-1)d)} \]
Setting this equal to \( \frac{m^2}{n^2} \):
\[ \frac{m(2a + (m-1)d)}{n(2a + (n-1)d)} = \frac{m^2}{n^2} \]

5. **Cross Multiplying**:
Cross multiplying gives:
\[ m(2a + (m-1)d) \cdot n^2 = n(2a + (n-1)d) \cdot m^2 \]

6. **Analyzing the Statements**:
- **Statement 1**: \( d = 2a \)
- **Statement 2**: \( \frac{a_m}{a_n} = \frac{2m+1}{2n+1} \)

7. **Finding \( a_m \) and \( a_n \)**:
The \( m \)-th term of the AP is given by:
\[ a_m = a + (m-1)d \]
The \( n \)-th term is:
\[ a_n = a + (n-1)d \]

8. **Substituting for \( d \)**:
If we assume \( d = 2a \), then:
\[ a_m = a + (m-1)(2a) = a + 2am - 2a = 2am - a \]
\[ a_n = a + (n-1)(2a) = a + 2an - 2a = 2an - a \]

9. **Finding the Ratio**:
Now, substituting these into the ratio:
\[ \frac{a_m}{a_n} = \frac{2am - a}{2an - a} = \frac{2m - 1}{2n - 1} \]
This does not match Statement 2, which claims:
\[ \frac{a_m}{a_n} = \frac{2m + 1}{2n + 1} \]
Therefore, Statement 2 is false.

10. **Conclusion**:
Since we have shown that \( d = 2a \) holds true, Statement 1 is true, and Statement 2 is false.

### Final Answer:
- **Statement 1**: True
- **Statement 2**: False