Suppose l, m, n respectively represent the co | 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 $\left(a x^2+\frac{b}{x^3}\right)^{15}$ is given by $T_{r+1} = {}^{15}C_r (a x^2)^{15-r} (b x^{-3})^r = {}^{1

Vedclass · Accepted Answer

$(9: 4)$

Vedclass · Answer

(B) The general term in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$ is given by $T_{r+1} = {}^{15}C_r (a x^2)^{15-r} (b x^{-3})^r = {}^{15}C_r a^{15-r} b^r x^{30-5r}$.For $x^{10}$,$30-5r = 10$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$. Thus,$l = {}^{15}C_4 a^{11} b^4$.For the constant term,$30-5r = 0 \Rightarrow r = 6$. Thus,$m = {}^{15}C_6 a^9 b^6$.For $x^{-10}$,$30-5r = -10$ $\Rightarrow 5r = 40$ $\Rightarrow r = 8$. Thus,$n = {}^{15}C_8 a^7 b^8$.Given $\frac{l}{m} + \frac{m}{n} = \frac{26}{11}$.$\frac{l}{m} = \frac{{}^{15}C_4 a^{11} b^4}{{}^{15}C_6 a^9 b^6} = \frac{1365}{5005} \cdot \frac{a^2}{b^2} = \frac{3}{11} \cdot \frac{a^2}{b^2}$.$\frac{m}{n} = \frac{{}^{15}C_6 a^9 b^6}{{}^{15}C_8 a^7 b^8} = \frac{5005}{6435} \cdot \frac{a^2}{b^2} = \frac{7}{9} \cdot \frac{a^2}{b^2}$.Substituting these into the equation: $\frac{a^2}{b^2} \left( \frac{3}{11} + \frac{7}{9} \right) = \frac{26}{11}$.$\frac{a^2}{b^2} \left( \frac{27 + 77}{99} \right) = \frac{26}{11}$ $\Rightarrow \frac{a^2}{b^2} \left( \frac{104}{99} \right) = \frac{26}{11}$.$\frac{a^2}{b^2} = \frac{26}{11} \cdot \frac{99}{104} = \frac{9}{4}$.Therefore,$a^2 : b^2 = 9 : 4$.

Suppose $l, m, n$ respectively represent the coefficient of $x^{10}$,the constant term,and the coefficient of $x^{-10}$ in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$. If $\frac{l}{m}+\frac{m}{n}=\frac{26}{11}$,then $a^2: b^2=$

Similar Questions

If the value of $C_{0}+2 \cdot C_{1}+3 \cdot C_{2}+\ldots+(n+1) \cdot C_{n}=576$,then $n$ is equal to

If the coefficient of ${x^7}$ in ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$ is equal to the coefficient of ${x^{ - 7}}$ in ${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$,then $ab =$

The coefficient of the term independent of $x$ in the expansion of $(1 + x + 2x^3) \left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is

If $n$ is a positive integer and $C_k = {^nC_k}$,then the value of $\sum\limits_{k = 1}^n {k^3\left( {\frac{C_k}{C_{k - 1}}} \right)^2}$ is:

The second,third and fourth terms in the binomial expansion $(x+a)^n$ are $240, 720$ and $1080$ respectively. Find $x, a$ and $n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module