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If $^{n} C_{4},^{n} C_{5}$ and $^{n} C_{6}$ are in A.P., then a value of $n$ can be

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Correct option is B

As nC4,nC5 and nC6 are in A.P.2 nC5=nC4+nC6⇒ 2= nC4 nC5+ nC6 nC5=n!4!(n−4)!5!(n−5)!n!+n!6!(n−6)!5!(n−5)!n!⇒ 2=5n−4+n−56⇒n2−21n+98=0⇒ n=7,14.

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If nC4,nC5 and nC6 are in A.P., then a value of n can be

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If $^{n} C_{4},^{n} C_{5}$ and $^{n} C_{6}$ are in A.P., then a value of $n$ can be