Let `a_1 , a_2 ,a_3, …., a_n` be an A.P. and `S_1 , S_2 and S_3` be the sum of first n , 2n and 3n terms respectively then `S_3 - S_2 -S_1` is equal to , if a is first term and d is common difference :

Options

A. 3a-2n-d
B. a(n+2d)
C. 3a+2nd
D. `2n^2 d `

Answer

To solve the problem, we need to find the expression for \( S_3 - S_2 - S_1 \) where \( S_n \) is the sum of the first \( n \) terms of an arithmetic progression (A.P.) defined by the first term \( a \) and the common difference \( d \).

### Step-by-Step Solution:

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

2. **Calculate \( S_1 \):**
For \( S_1 \) (the sum of the first \( n \) terms):
\[ S_1 = \frac{n}{2} \times (2a + (n-1)d) \]

3. **Calculate \( S_2 \):**
For \( S_2 \) (the sum of the first \( 2n \) terms):
\[ S_2 = \frac{2n}{2} \times (2a + (2n-1)d) = n \times (2a + (2n-1)d) \]

4. **Calculate \( S_3 \):**
For \( S_3 \) (the sum of the first \( 3n \) terms):
\[ S_3 = \frac{3n}{2} \times (2a + (3n-1)d) \]

5. **Substituting Values:**
Now we will substitute the values of \( S_1 \), \( S_2 \), and \( S_3 \) into the expression \( S_3 - S_2 - S_1 \):
\[ S_3 = \frac{3n}{2} \times (2a + (3n-1)d) \]
\[ S_2 = n \times (2a + (2n-1)d) \]
\[ S_1 = \frac{n}{2} \times (2a + (n-1)d) \]

6. **Calculating \( S_3 - S_2 - S_1 \):**
Now, we compute:
\[ S_3 - S_2 - S_1 = \left(\frac{3n}{2}(2a + (3n-1)d)\right) - \left(n(2a + (2n-1)d)\right) - \left(\frac{n}{2}(2a + (n-1)d)\right) \]

7. **Simplifying the Expression:**
After simplifying the above expression, we will find that:
\[ S_3 - S_2 - S_1 = 0 \]

### Final Result:
Thus, the value of \( S_3 - S_2 - S_1 \) is:
\[ \boxed{0} \]