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 statements about the arithmetic progression (AP) defined by its first term \( a \) and common difference \( d \).

### Step 1: Understanding the Sum of n Terms of an AP
The sum of the first \( n \) terms of an AP is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
This can be simplified to:
\[ S_n = \frac{n}{2} \left(2a + nd - d\right) = \frac{n}{2} \left((n-1)d + 2a\right) \]

### Step 2: Setting Up the Given Ratio
We are given that:
\[ \frac{S_m}{S_n} = \frac{m^2}{n^2} \]
Substituting the formula for \( S_m \) and \( S_n \):
\[ \frac{\frac{m}{2} \left(2a + (m-1)d\right)}{\frac{n}{2} \left(2a + (n-1)d\right)} = \frac{m^2}{n^2} \]
This simplifies to:
\[ \frac{m(2a + (m-1)d)}{n(2a + (n-1)d)} = \frac{m^2}{n^2} \]

### Step 3: Cross Multiplying
Cross multiplying gives us:
\[ m(2a + (m-1)d) \cdot n^2 = n(2a + (n-1)d) \cdot m^2 \]
Expanding both sides:
\[ mn^2(2a + (m-1)d) = nm^2(2a + (n-1)d) \]
Dividing both sides by \( mn \) (assuming \( m, n \neq 0 \)):
\[ n(2a + (m-1)d) = m(2a + (n-1)d) \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ 2an - 2am + n(m-1)d = m(n-1)d \]
This simplifies to:
\[ 2a(n - m) = d(m(n - 1) - n(m - 1)) \]

### Step 5: Analyzing the Statements
1. **Statement 1: \( d = 2a \)**
- If we assume \( d = 2a \), substituting this into our equation will help us check if it holds true.
- If \( d = 2a \), substituting gives us a consistent equation.

2. **Statement 2: \( \frac{a_m}{a_n} = \frac{2m + 1}{2n + 1} \)**
- The nth term of an AP is given by:
\[ a_n = a + (n-1)d \]
- Thus, we have:
\[ a_m = a + (m-1)d \quad \text{and} \quad a_n = a + (n-1)d \]
- Substituting \( d = 2a \) into \( a_m \) and \( a_n \) gives:
\[ a_m = a + (m-1)(2a) = a + 2a(m-1) = 2am - a \]
\[ a_n = a + (n-1)(2a) = a + 2a(n-1) = 2an - a \]
- Thus:
\[ \frac{a_m}{a_n} = \frac{2am - a}{2an - a} = \frac{2m - 1}{2n - 1} \]
- This does not match \( \frac{2m + 1}{2n + 1} \).

### Conclusion
- **Statement 1 is true:** \( d = 2a \).
- **Statement 2 is false:** \( \frac{a_m}{a_n} \neq \frac{2m + 1}{2n + 1} \).