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Question: If \(nC_{4},\mspace{6mu}^{n}C_{5}\) and \(nC_{6}\) are in A.P., then n =...

If nC4, 6munC5nC_{4},\mspace{6mu}^{n}C_{5} and nC6nC_{6} are in A.P., then n =

A

7 or 14

B

7

C

14

D

None of these

Answer

7 or 14

Explanation

Solution

Given nC4, 6munC5, 6munC6nC_{4},\mspace{6mu}^{n}C_{5},\mspace{6mu}^{n}C_{6} are in A.P.

⇒ 2 6mux 6munC5= 6munC4+nC62\mspace{6mu} x\mspace{6mu}^{n}C_{5} = \mspace{6mu}^{n}C_{4} +^{n}C_{6}

⇒2∠n∠n−5 6mu∠5 6mu= 6mu∠n∠n−4 6mu∠4+∠n∠n−6 6mu∠6\frac{2\angle n}{\angle n - 5\mspace{6mu}\angle 5}\mspace{6mu} = \mspace{6mu}\frac{\angle n}{\angle n - 4\mspace{6mu}\angle 4} + \frac{\angle n}{\angle n - 6\mspace{6mu}\angle 6}⇒ 2∠n−5 6mu∠5 6mu= 6mu1∠n−4 6mu∠4 6mu+ 6mu1∠n−6 6mu∠6\frac{2}{\angle n - 5\mspace{6mu}\angle 5}\mspace{6mu} = \mspace{6mu}\frac{1}{\angle n - 4\mspace{6mu}\angle 4}\mspace{6mu} + \mspace{6mu}\frac{1}{\angle n - 6\mspace{6mu}\angle 6}Multiplying throughout by ∠n−4\angle n - 4 ∠6\angle 6, we get

2 x (n – 4) x 6 = 6 x 5 + (n – 4) (n – 5)

⇒ n2 - 21ln + 98 = 0

⇒ (n – 7) (n – 14) = 0

⇒ n = 7 or 14.