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Q.

Let S₁be the sum of first 2n terms of an arithmetic progression. Let S₂ be the sum of first 4n terms of the same arithmetic progression. If (S₂ – S₁) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

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a

5000

b

7000

c

1000

d

3000

answer is D.

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Detailed Solution

$S_{1} = \frac{2 n}{2} [2 a + (2 n - 1) d] S_{2} = \frac{4 n}{2} [2 a + (4 n - 1) d] S_{2} - S_{1} = 1000 \Rightarrow n [4 a + (8 n - 2) d - 2 a - (2 n - 1) d] = 1000 \Rightarrow n [2 a + (6 n - 1) d] = 1000 S_{6 n} = \frac{6 n}{2} [2 a + (6 n - 1) d] = 3 (1000) = 3000$

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