If `S_1`is the sum of first 2n terms and `S_2` is the sum of first 4n term and `S_2-S_1` is equal to 1000 then `S_3` sum of first 6n terms of same A.P

Options

A. 3000
B. 5000
C. 7000
D. 9000

Answer

To solve the problem, we need to find the sum \( S_3 \) of the first \( 6n \) terms of an arithmetic progression (A.P.) given that \( S_2 - S_1 = 1000 \).

### Step-by-Step Solution:

1. **Understanding the Sums**:
- The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
- Here, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

2. **Calculate \( S_1 \) and \( S_2 \)**:
- For \( S_1 \) (sum of first \( 2n \) terms):
\[ S_1 = S_{2n} = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]
- For \( S_2 \) (sum of first \( 4n \) terms):
\[ S_2 = S_{4n} = \frac{4n}{2} \left(2a + (4n-1)d\right) = 2n \left(2a + (4n-1)d\right) \]

3. **Calculate \( S_2 - S_1 \)**:
- We know from the problem that \( S_2 - S_1 = 1000 \):
\[ S_2 - S_1 = 2n \left(2a + (4n-1)d\right) - n \left(2a + (2n-1)d\right) \]
- Simplifying this:
\[ = 2n(2a + 4nd - d) - n(2a + 2nd - d) \]
\[ = 2n(2a + 4nd - d) - n(2a + 2nd - d) \]
\[ = 2n(2a + 4nd - d) - n(2a + 2nd - d) \]
\[ = n \left[ 2(2a + 4nd - d) - (2a + 2nd - d) \right] \]
\[ = n \left[ 4a + 8nd - 2d - 2a - 2nd + d \right] \]
\[ = n \left[ 2a + 6nd - d \right] \]
- Setting this equal to 1000:
\[ n(2a + 6nd - d) = 1000 \]

4. **Calculate \( S_3 \)**:
- Now, we need to find \( S_3 \) (sum of first \( 6n \) terms):
\[ S_3 = S_{6n} = \frac{6n}{2} \left(2a + (6n-1)d\right) = 3n \left(2a + (6n-1)d\right) \]
- We can express \( S_3 \) in terms of the previous equation:
\[ S_3 = 3n \left(2a + 6nd - d\right) \]

5. **Relate \( S_3 \) to \( S_2 - S_1 \)**:
- From the equation \( n(2a + 6nd - d) = 1000 \):
\[ S_3 = 3 \cdot 1000 = 3000 \]

### Final Answer:
\[ S_3 = 3000 \]