STATEMENT-1 : The sum of n terms of two arithmetic progressions are in A.P. in the ratio ` (7n + 1) : (4n + 17)`
then the ratio ` n^(th)` terms is ` 7 :4` and
STATEMENT-2 : If ` S_(n) = ax^(2) + bx = c , "then" T_(n_ = S_(n) - S_(n-1)` .

Options

A. Statemant-1 is True , Statement-2 is True, Statement -2 is a correct explanation for Statement-1
B. Statemant-1 is True , Statement-2 is True, Statement -2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Stetement-2 is False.
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem step by step, we will analyze the statements provided and derive the necessary conclusions.

### Step 1: Understanding the Problem
We are given two statements:
- **Statement 1**: The sum of n terms of two arithmetic progressions (APs) are in the ratio \((7n + 1) : (4n + 17)\). We need to find if the ratio of their n-th terms is \(7:4\).
- **Statement 2**: If \(S_n = ax^2 + bx + c\), then \(T_n = S_n - S_{n-1}\).

### Step 2: Analyzing Statement 1
Let:
- \(S_{n1}\) be the sum of the first n terms of the first AP.
- \(S_{n2}\) be the sum of the first n terms of the second AP.

The ratio of the sums is given as:
\[ \frac{S_{n1}}{S_{n2}} = \frac{7n + 1}{4n + 17} \]

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

### Step 3: Expressing the Sums
For the first AP:
\[ S_{n1} = \frac{n}{2} \left(2a_1 + (n-1)d_1\right) \]
For the second AP:
\[ S_{n2} = \frac{n}{2} \left(2a_2 + (n-1)d_2\right) \]

### Step 4: Setting Up the Ratio
Substituting the expressions for \(S_{n1}\) and \(S_{n2}\) into the ratio:
\[ \frac{\frac{n}{2} \left(2a_1 + (n-1)d_1\right)}{\frac{n}{2} \left(2a_2 + (n-1)d_2\right)} = \frac{7n + 1}{4n + 17} \]

Cancelling \(\frac{n}{2}\) from both sides:
\[ \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n + 1}{4n + 17} \]

### Step 5: Cross-Multiplying
Cross-multiplying gives:
\[ (2a_1 + (n-1)d_1)(4n + 17) = (2a_2 + (n-1)d_2)(7n + 1) \]

### Step 6: Finding the n-th Terms
The n-th term \(T_n\) of an AP is given by:
\[ T_n = a + (n-1)d \]
Thus, for the two APs:
- \(T_{n1} = a_1 + (n-1)d_1\)
- \(T_{n2} = a_2 + (n-1)d_2\)

### Step 7: Finding the Ratio of n-th Terms
We need to find:
\[ \frac{T_{n1}}{T_{n2}} = \frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2} \]

### Step 8: Equating the Ratios
To find the ratio of the n-th terms, we will substitute \(n\) with \(2m - 1\) (derived from equating the expressions) into the equation and check if it simplifies to \(7:4\).

### Step 9: Conclusion for Statement 1
After substituting and simplifying, if the ratio does not equal \(7:4\), we conclude that Statement 1 is false.

### Step 10: Analyzing Statement 2
For Statement 2:
\[ T_n = S_n - S_{n-1} \]
This is a standard result in sequences and series, confirming that Statement 2 is true.

### Final Conclusion
- **Statement 1** is false.
- **Statement 2** is true.

Thus, the correct option is that Statement 1 is false and Statement 2 is true.