If the second,third,and fourth terms in the e | vedclass

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

Question

(D) The general term in the expansion of $(x + a)^n$ is $T_{r+1} = {}^nC_r x^{n-r} a^r$.Given $T_2 = {}^nC_1 x^{n-1} a = 240$ $(i)$,$T_3 = {}^nC_2 x^{

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The general term in the expansion of $(x + a)^n$ is $T_{r+1} = {}^nC_r x^{n-r} a^r$.Given $T_2 = {}^nC_1 x^{n-1} a = 240$ $(i)$,$T_3 = {}^nC_2 x^{n-2} a^2 = 720$ $(ii)$,$T_4 = {}^nC_3 x^{n-3} a^3 = 1080$ $(iii)$.Dividing $(ii)$ by $(i)$,we get $\frac{T_3}{T_2} = \frac{{}^nC_2 x^{n-2} a^2}{{}^nC_1 x^{n-1} a} = \frac{n-1}{2} \cdot \frac{a}{x} = \frac{720}{240} = 3$ $\Rightarrow \frac{a}{x} = \frac{6}{n-1}$.Dividing $(iii)$ by $(ii)$,we get $\frac{T_4}{T_3} = \frac{{}^nC_3 x^{n-3} a^3}{{}^nC_2 x^{n-2} a^2} = \frac{n-2}{3} \cdot \frac{a}{x} = \frac{1080}{720} = \frac{3}{2}$ $\Rightarrow \frac{a}{x} = \frac{9}{2(n-2)}$.Equating the two expressions for $\frac{a}{x}$:$\frac{6}{n-1} = \frac{9}{2(n-2)}$$12(n-2) = 9(n-1)$$12n - 24 = 9n - 9$$3n = 15 \Rightarrow n = 5$.

If the second,third,and fourth terms in the expansion of $(x + a)^n$ are $240, 720,$ and $1080$ respectively,then the value of $n$ is

Similar Questions

The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1:7:42$. Find $n$.

The number of rational terms in the expansion of $(3^{1/8} + 5^{1/3})^{400}$ is

If the coefficients of the three successive terms in the binomial expansion of $(1 + x)^n$ are in the ratio $1 : 7 : 42,$ then the first of these terms in the expansion is (in $^{th}$)

If the $k^{\text{th}}$ term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is independent of $x$,then the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^k$ when $x = \frac{2}{3}$ is:

Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module