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:$

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$14$

Vedclass · Answer

(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:$2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$Dividing both sides by $n!$ and multiplying by $6!(n-4)!$:$2 \cdot \frac{6 \cdot (n-4)}{5} = 6 \cdot 5 + (n-4)(n-5)$$12(n-4) = 30 + n^2 - 9n + 20$$12n - 48 = n^2 - 9n + 50$$n^2 - 21n + 98 = 0$$(n-7)(n-14) = 0$Thus,$n = 7$ or $n = 14$. Since $n \ge 6$ is required for $^nC_6$ to be defined,both are possible,but $14$ is given in the options.

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

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