Let a and b be two nonzero real numbers. If t | vedclass

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

Question

(B) The general term in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is $T_{r+1} = {}^4C_r (ax^2)^{4-r} (\frac{70}{27bx})^r = {}^4C_r a^{4-r} (\frac{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The general term in the expansion of $(ax^2 + \frac{70}{27bx})^4$ is $T_{r+1} = {}^4C_r (ax^2)^{4-r} (\frac{70}{27bx})^r = {}^4C_r a^{4-r} (\frac{70}{27b})^r x^{8-3r}$.For the coefficient of $x^5$,we set $8-3r = 5$,which gives $r=1$.The coefficient is ${}^4C_1 a^3 (\frac{70}{27b}) = 4a^3 \cdot \frac{70}{27b} = \frac{280a^3}{27b}$.The general term in the expansion of $(ax - \frac{1}{bx^2})^7$ is $T_{r+1} = {}^7C_r (ax)^{7-r} (-\frac{1}{bx^2})^r = {}^7C_r a^{7-r} (-\frac{1}{b})^r x^{7-3r}$.For the coefficient of $x^{-5}$,we set $7-3r = -5$,which gives $3r = 12$,so $r=4$.The coefficient is ${}^7C_4 a^3 (-\frac{1}{b})^4 = 35 \cdot \frac{a^3}{b^4} = \frac{35a^3}{b^4}$.Equating the two coefficients: $\frac{35a^3}{b^4} = \frac{280a^3}{27b}$.Dividing by $35a^3$ (since $a 
eq 0$),we get $\frac{1}{b^3} = \frac{8}{27}$.Thus,$b^3 = \frac{27}{8}$,which implies $b = \frac{3}{2}$.Therefore,$2b = 2(\frac{3}{2}) = 3$.

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

Similar Questions

If the coefficient of $x^7$ in $(ax - \frac{1}{bx^2})^{13}$ and the coefficient of $x^{-5}$ in $(ax + \frac{1}{bx^2})^{13}$ are equal,then $a^4 b^4$ is equal to :

The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

If the coefficients of $r$-th and $(r+1)$-th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

Let $(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$. Then $\frac{a_7}{a_{13}}$ is equal to

If the second term of the expansion $\left[ a^{\frac{1}{13}} + \frac{a}{\sqrt{a^{-1}}} \right]^n$ is $14a^{5/2}$,then the value of $\frac{^nC_3}{^nC_2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module