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Q.

If $^{2 n} C_{4},^{2 n} C_{5}$ and $^{2 n} C_{6}$ are in A.P, then n is equal to

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a

14

b

12

c

7

d

6

answer is C.

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Detailed Solution

$\begin{array}{l} 2 (^{2 n} C_{5}) =^{2 n} C_{4} +^{2 n} C_{6} \\ \Rightarrow 2 = \frac{^{2 n} C_{4}}{^{2 n} C_{5}} + \frac{^{2 n} C_{6}}{2 n} = \frac{5}{2 n - 4} + \frac{2 n - 5}{6} \\ \Rightarrow 12 (2 n - 4) = 30 + (2 n - 4) (2 n - 5) \\ \Rightarrow 2 n^{2} - 21 n + 49 = 0 \Rightarrow n = 7 \end{array}$

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