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Question

Let Sn denote the sum of the first n terms of an A.P. If S2n=3Sn, then S3nSn=

A
3
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B
6
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C
7
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D
5
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Solution

The correct option is A 6
Let a be the first term and d the common difference of the given A.P. Then,
S2n=3Sn
2n2[a+(2n1)d]=3n2[a+(n1)d]

2[a+(2n1)d]=3[a+(n1)d]
2a(3n34n+2)d=0
2a(n+1)d=0
2a=(n+1)d
Now, S3nSn=3n2[2a+(3n1)d]n2[2a+(n1)d]

=3[(n+1)d+(3n1)d][(n+1)d+(n1)d]

=12nd2nd=6

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