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UPSC NDA Math Model Paper 7

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Question : 116 of 120

Marks: +1, -0

If

{}^{n}C_{4}, {}^{n}C_{5}

{}^{n}C_{6}

are in AP, then

n

can be

9
11
14
12

Solution:

If

{}^{n}C_{4}, {}^{n}C_{5}

and

{}^{n}C_{6}

are in AP, then

2^{n}C_{5} = {}^{n}C_{4} + {}^{n}C_{6}

[If

a, b, c

are in

AP

, then

2b = a + c]

\Rightarrow \frac{2 n!}{5! (n-5)!} = \frac{n!}{4! (n-4)!} + \frac{n!}{6! (n-6)!}

\left[ \because {}^{n}C_{r} = \frac{n!}{r! (n-r)!} \right]

\Rightarrow \frac{2}{5! (n-5) (n-6)!}

= \frac{1}{4! (n-4) (n-5) (n-6)!} + \frac{1}{6.5.4! (n-6)!}

\Rightarrow \frac{2}{5 (n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}

\Rightarrow \frac{2}{5 (n-5)} = \frac{30 + (n-4)(n-5)}{30 (n-4)(n-5)}

\Rightarrow 12(n-4) = 30 + n^{2} - 9n + 20

\Rightarrow n^{2} - 21n + 98 = 0

\Rightarrow n^{2} - 14n - 7n + 98 = 0

\Rightarrow n(n-14) - 7(n-14) = 0

\Rightarrow (n-7)(n-14) = 0

\Rightarrow n = 7

or 14

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