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Question: If \(5Cn,\mspace{6mu} 2\mspace{6mu}^{4}C_{n}\mspace{6mu} and\mspace{6mu}^{6}C_{n}\mspace{6mu}\) are ...

If 5Cn, 6mu2 6mu4Cn 6muand 6mu6Cn 6mu5Cn,\mspace{6mu} 2\mspace{6mu}^{4}C_{n}\mspace{6mu} and\mspace{6mu}^{6}C_{n}\mspace{6mu} are in H.P., then n =

A

2

B

15

C

3

D

None of these

Answer

2

Explanation

Solution

Given 15Cn, 6mu124Cn, 6mu16Cn\frac{1}{5C_{n}},\mspace{6mu}\frac{1}{2^{4}C_{n}},\mspace{6mu}\frac{1}{6C_{n}} are in A.P.

⇒ 22 6mu4Cn=15Cn+16Cn\frac{2}{2^{\mspace{6mu} 4}C_{n}} = \frac{1}{5C_{n}} + \frac{1}{6C_{n}}

⇒ ∠4−n 6mu∠n∠4 6mu= 6mu∠5−n 6mu∠n∠5 6mu+ 6mu∠6−n 6mu∠n∠6\frac{\angle 4 - n\mspace{6mu}\angle n}{\angle 4}\mspace{6mu} = \mspace{6mu}\frac{\angle 5 - n\mspace{6mu}\angle n}{\angle 5}\mspace{6mu} + \mspace{6mu}\frac{\angle 6 - n\mspace{6mu}\angle n}{\angle 6}

⇒ ∠4−n∠4 6mu= 6mu∠5−n∠5 6mu+ 6mu∠6−n∠6\frac{\angle 4 - n}{\angle 4}\mspace{6mu} = \mspace{6mu}\frac{\angle 5 - n}{\angle 5}\mspace{6mu} + \mspace{6mu}\frac{\angle 6 - n}{\angle 6}

Multiplying throughout by ∠6∠4−n\frac{\angle 6}{\angle 4 - n}

6 x 5 = 6 x (5 – n) + (6 – n) (5 – n)

⇒ n2- 17n + 30 = 0 ⇒ n = 2, 15.

For n = 15, 4Cn4C_{n} is meaningless, therefore, n = 2 is the only solution.