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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

1000

c

3000

d

7000

answer is D.

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Detailed Solution

S1=2n22a+(2n-1)d S2=4n22a+(4n-1)d S2-S1=1000 ⇒  n4a+(8n-2)d-2a-(2n-1)d=1000 ⇒  n2a+(6n-1)d=1000 S6n=6n22a+(6n-1)d=3(1000)=3000

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