If `S_n` denotes the sum of n terms of an A.P. whose common difference is d and first term is a, find `S_n-2S_(n-1)+S_(n-2)`

Options

A. `d=S_(n)-S_(n-1)+S_(n-1)`
B. `d=S_(n)-2S_(n-1)-S_(n-2)`
C. `d=S_(n)-2S_(n-1)+S_(n-2)`
D. none of these

Answer

To solve the problem, we need to find the expression \( S_n - 2S_{n-1} + S_{n-2} \) where \( S_n \) is the sum of the first \( n \) terms of an arithmetic progression (A.P.) with first term \( a \) and common difference \( d \).

### Step-by-step Solution:

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

2. **Calculate \( S_{n-1} \)**:
Using the formula for \( S_n \), we can find \( S_{n-1} \):
\[ S_{n-1} = \frac{n-1}{2} \left( 2a + (n-2)d \right) \]

3. **Calculate \( S_{n-2} \)**:
Similarly, we find \( S_{n-2} \):
\[ S_{n-2} = \frac{n-2}{2} \left( 2a + (n-3)d \right) \]

4. **Substituting into the Expression**:
Now we substitute \( S_n \), \( S_{n-1} \), and \( S_{n-2} \) into the expression \( S_n - 2S_{n-1} + S_{n-2} \):
\[ S_n - 2S_{n-1} + S_{n-2} = \left( \frac{n}{2} \left( 2a + (n-1)d \right) \right) - 2 \left( \frac{n-1}{2} \left( 2a + (n-2)d \right) \right) + \left( \frac{n-2}{2} \left( 2a + (n-3)d \right) \right) \]

5. **Simplifying the Expression**:
Let's simplify this step by step:
- First, simplify \( -2S_{n-1} \):
\[ -2S_{n-1} = - (n-1) \left( 2a + (n-2)d \right) \]
- Now combine all three terms:
\[ S_n - 2S_{n-1} + S_{n-2} = \frac{n}{2} \left( 2a + (n-1)d \right) - (n-1) \left( 2a + (n-2)d \right) + \frac{n-2}{2} \left( 2a + (n-3)d \right) \]

6. **Final Result**:
After simplifying the expression, we find that:
\[ S_n - 2S_{n-1} + S_{n-2} = d \]

### Conclusion:
Thus, the final answer is:
\[ S_n - 2S_{n-1} + S_{n-2} = d \]