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Question

If the coefficients of 5th,6th and 7th terms in the expansion of (1+x)n are in A.P., then the value n is equal to

A
7 only
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B
14 only
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C
7 or 14
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D
None of these
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Solution

The correct option is C 7 or 14
Here, coefficient of T5 will be nC4, that of T6 will be nC5 and that of T7 will be nC6.
Since, it is given they are in A.P.
2nC5= nC4+nC6
Hence,
2[n!(n5)!5!]=[n!(n4)!4!+n!(n6)!6!]
2[1(n5)5]=[1(n4)(n5)+16×5]
12n48=30+n29n+20n221n+98=0
n=7 or n=14

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