Let the coefficients of the middle terms in t | vedclass

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

Question

(A) The middle term of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4}$ is the $3^{rd}$ term: $T_{3} = {}^{4}C_{2} \left(\frac{1}{\sqrt{6}}\right)^{2} (\

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(A) The middle term of $\left(\frac{1}{\sqrt{6}}+\beta xight)^{4}$ is the $3^{rd}$ term: $T_{3} = {}^{4}C_{2} \left(\frac{1}{\sqrt{6}}ight)^{2} (\beta x)^{2} = 6 	imes \frac{1}{6} \beta^{2} x^{2} = \beta^{2} x^{2}$. Coefficient is $\beta^{2}$.The middle term of $(1-3 \beta x)^{2}$ is the $2^{nd}$ term: $T_{2} = {}^{2}C_{1} (1)^{1} (-3 \beta x)^{1} = -6 \beta x$. Coefficient is $-6 \beta$.The middle term of $\left(1-\frac{\beta}{2} xight)^{6}$ is the $4^{th}$ term: $T_{4} = {}^{6}C_{3} (1)^{3} \left(-\frac{\beta}{2} xight)^{3} = 20 	imes \left(-\frac{\beta^{3}}{8}ight) x^{3} = -\frac{5}{2} \beta^{3} x^{3}$. Coefficient is $-\frac{5}{2} \beta^{3}$.Since these are in $A.P.$,$2(-6 \beta) = \beta^{2} - \frac{5}{2} \beta^{3}$.$-12 \beta = \beta^{2} - \frac{5}{2} \beta^{3} \implies \frac{5}{2} \beta^{2} - \beta - 12 = 0 \implies 5 \beta^{2} - 2 \beta - 24 = 0$.Solving $5 \beta^{2} - 12 \beta + 10 \beta - 24 = 0 \implies \beta(5 \beta - 12) + 2(5 \beta - 12) = 0 \implies (\beta + 2)(5 \beta - 12) = 0$.Since $\beta > 0$,$\beta = \frac{12}{5}$.The terms are $\beta^{2}, -6 \beta, -\frac{5}{2} \beta^{3}$.$d = -6 \beta - \beta^{2} = -6 \left(\frac{12}{5}ight) - \left(\frac{12}{5}ight)^{2} = -\frac{72}{5} - \frac{144}{25} = \frac{-360 - 144}{25} = -\frac{504}{25}$.$50 - \frac{2d}{\beta^{2}} = 50 - \frac{2(-504/25)}{(12/5)^{2}} = 50 - \frac{-1008/25}{144/25} = 50 + \frac{1008}{144} = 50 + 7 = 57$.

Similar Questions

If the coefficients of the second,third,and fourth terms in the expansion of $(1 + x)^{2n}$ are in $A.P.$,then $2n^2 - 9n + 7$ is equal to

In the expansion of $(1 + x)^{43}$,if the coefficients of the $(2r + 1)^{th}$ and the $(r + 2)^{th}$ terms are equal,the value of $r$ is:

What is the constant term in the binomial expansion of $(1+3x)^n \left(1+\frac{1}{3x}\right)^n$?

The numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$ is

In the expansion of ${\left( \frac{a}{x} + bx \right)^{12}}$,the coefficient of $x^{-10}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module