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Question

If the coefficient of 2nd, 3rd & 4th terms in the expansion (1+x)2n are in A.P., which of the following hold good

A
n29n+7=0
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B
2n29n7=0
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C
2n29n+7=0
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D
9n27n+2=0
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Solution

The correct option is C 2n29n+7=0
Hence
2nC1, 2nC2 and 2nC3 are in A.P.
Therefore
It must follow that
(2n2(2))2=(2n+2)
(2n4)2=2(n+1)
4(n2)2=2(n+1)
2(n24n+8)=n+1
2n28n+8=n+1
2n29n+7=0

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