If `S_(n)` denote the sum to n terms of an A.P. whose
first term is a and common differnece is d , then
`S_(n) - 2S_(n-1) + S_(n-2)` is equal to

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 the first term \( a \) and common difference \( d \).

### Step-by-step Solution:

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

2. **Calculate \( S_{n-1} \)**:
Substitute \( n-1 \) into the formula:
\[ S_{n-1} = \frac{n-1}{2} \left( 2a + (n-2)d \right) \]

3. **Calculate \( S_{n-2} \)**:
Substitute \( n-2 \) into the formula:
\[ S_{n-2} = \frac{n-2}{2} \left( 2a + (n-3)d \right) \]

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

5. **Simplifying the expression**:
Let's simplify each term:
- The first term is:
\[ \frac{n}{2} \left( 2a + (n-1)d \right) \]
- The second term becomes:
\[ - (n-1) \left( 2a + (n-2)d \right) \]
- The third term is:
\[ \frac{n-2}{2} \left( 2a + (n-3)d \right) \]

Now, combine these terms:
\[ S_n - 2S_{n-1} + S_{n-2} = \frac{n}{2} (2a + (n-1)d) - (n-1)(2a + (n-2)d) + \frac{n-2}{2} (2a + (n-3)d) \]

6. **Final simplification**:
After performing the algebraic simplifications, we find that:
\[ S_n - 2S_{n-1} + S_{n-2} = d \]

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