Mathematics - Binomial Theorem Question with Solution | TestHub

MathematicsBinomial TheoremProperties of Binomial coefficientsEasy2 minPYQ_2019

MathematicsEasysingle choice

If $^{n} C_{4},^{n} C_{5}$ and $^{n} C_{6}$ are in A.P., then $n$ can be

Answer:

Solution:

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

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

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

$\Rightarrow \frac{2}{5! (n - 5)!} = \frac{30 + (n - 5) (n - 4)}{6! (n - 4)!}$

$\Rightarrow 6 \times 2 (n - 4) = 30 + n^{2} - 9 n + 20$

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

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

$\Rightarrow n = 7$ or $14$

Stream:JEESubject:MathematicsTopic:Binomial TheoremSubtopic:Properties of Binomial coefficients

⏱ 2mℹ️ Source: PYQ_2019

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