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If S1is the sum of first 2n terms and S2...

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

A

3000

B

5000

C

7000

D

9000

Text Solution

AI Generated Solution

The correct Answer is:
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: \[ S_3 = 3000 \]
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Knowledge Check

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