Let the coefficients of three consecutive ter | vedclass

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

Question

(C) Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$. Their coefficients are $^nC_{r-1} 2^{r-1}, ^nC_r 2^r, ^nC_{r+1} 2^{r+1}$.Given the rat

Vedclass · Accepted Answer

$1120$

Vedclass · Answer

(C) Let the three consecutive terms be $T_r, T_{r+1}, T_{r+2}$. Their coefficients are $^nC_{r-1} 2^{r-1}, ^nC_r 2^r, ^nC_{r+1} 2^{r+1}$.Given the ratio $^nC_{r-1} 2^{r-1} : ^nC_r 2^r : ^nC_{r+1} 2^{r+1} = 2 : 5 : 8$.From $\frac{^nC_{r-1} 2^{r-1}}{^nC_r 2^r} = \frac{2}{5}$,we get $\frac{r}{n-r+1} 	imes \frac{1}{2} = \frac{2}{5}$ $\Rightarrow \frac{r}{n-r+1} = \frac{4}{5}$ $\Rightarrow 5r = 4n - 4r + 4$ $\Rightarrow 9r - 4n = 4$ (Eq. $1$).From $\frac{^nC_r 2^r}{^nC_{r+1} 2^{r+1}} = \frac{5}{8}$,we get $\frac{r+1}{n-r} 	imes \frac{1}{2} = \frac{5}{8}$ $\Rightarrow \frac{r+1}{n-r} = \frac{5}{4}$ $\Rightarrow 4r + 4 = 5n - 5r$ $\Rightarrow 9r - 5n = -4$ (Eq. $2$).Subtracting Eq. $2$ from Eq. $1$: $(9r - 4n) - (9r - 5n) = 4 - (-4) \Rightarrow n = 8$.Substituting $n=8$ in Eq. $1$: $9r - 32 = 4$ $\Rightarrow 9r = 36$ $\Rightarrow r = 4$.The middle term coefficient is $^nC_r 2^r = ^8C_4 2^4 = 70 	imes 16 = 1120$.

Let the coefficients of three consecutive terms in the binomial expansion of $(1+2x)^n$ be in the ratio $2:5:8$. Then the coefficient of the term,which is in the middle of these three terms,is $...........$.

Similar Questions

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is equal to the coefficient of $x^{-5}$ in the expansion of $(ax - \frac{1}{bx^2})^7$,then the value of $2b$ is

If $n$ is the degree of the polynomial,$\left[ {\frac{1}{{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8 + \left[ {\frac{1}{{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8$ and $m$ is the coefficient of $x^{12}$ in it,then the ordered pair $(n, m)$ is equal to

The coefficient of $x^6$ in the binomial expansion of ${\left( \frac{4x^2}{3} - \frac{3}{2x} \right)^9}$ is

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $\frac{a}{b}$ is equal to

If the coefficient of the $4^{th}$ term in the expansion of $(a + b)^n$ is $56$,then $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module