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

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answer is B.

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Detailed Solution

As $^{n} C_{4},^{n} C_{5} {and}^{n} C_{6}$ are in A.P.

$2 (^{n} C_{5}) =^{n} C_{4} +^{n} C_{6}$

$\Rightarrow 2 = \frac{^{n} C_{4}}{^{n} C_{5}} + \frac{^{n} C_{6}}{^{n} C_{5}}$

$= \frac{n!}{4! (n - 4)!} \frac{5! (n - 5)!}{n!} + \frac{n!}{6! (n - 6)!} \frac{5! (n - 5)!}{n!}$

$\begin{array}{l} \Rightarrow 2 = \frac{5}{n - 4} + \frac{n - 5}{6} \Rightarrow n^{2} - 21 n + 98 = 0 \\ \Rightarrow n = 7, 14 . \end{array}$

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