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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:$

Vedclass · Accepted Answer

$14$

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:

Similar Questions

If $\omega$ is a non-real root of the equation $x^3 - 1 = 0$,then the value of $\sum_{r=1}^5 (1 + \omega^r + \omega^{2r})$ is

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

If ${x_1}, {x_2}, {x_3}$ as well as ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

The sum of the first $n$ natural numbers is:

How many terms of the $A.P.$ $1, 4, 7, \ldots$ are needed to give the sum $715$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module