If ^nC_4, ^nC_5, and ^nC_6 are in A.P., then | vedclass

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(B) Given that $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.$Therefore,$2(^nC_5) = ^nC_4 + ^nC_6$Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$2

Vedclass · Accepted Answer

$14$

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(B) Given that $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.$Therefore,$2(^nC_5) = ^nC_4 + ^nC_6$Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:$2 	imes \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$Dividing both sides by $n!$ and multiplying by $6!(n-4)!$:$\frac{2 	imes 6 	imes (n-4)}{5!} = \frac{6 	imes 5 	imes 4!}{4!} + \frac{(n-4)(n-5)}{6!/6!}$$\frac{12}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!}$Dividing by $\frac{1}{4!(n-6)!}$:$\frac{12}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}$Multiplying by $30(n-4)(n-5)$:$72(n-4) = 30 + (n-4)(n-5)$$72n - 288 = 30 + n^2 - 9n + 20$$n^2 - 81n + 338 = 0$$(n-14)(n-67) = 0$Since $n \ge 6$ for $^nC_6$ to be defined,$n = 14$ or $n = 67$. Among the options,$n = 14$ is correct.

If $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.,$ then $n$ can be

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