Question

In the expansion of ${(2–3x^3)}^{20}$, if the ratio of $10th$ term to $11th$ term is $\frac{45}{22}$, then $x$ is equal to

• A. $–\frac{2}{3}$
• B. $–\frac{3}{2}$
• C. ${\left(\frac{2}{3}\right)}^{-\frac{1}{3}}$
• D. ${\left(\frac{3}{2}\right)}^{-\frac{1}{3}}$

Answer 1

The correct option is A

$–\frac{2}{3}$

Explanation For The Correct Option:

Finding the value of $x$ using the given information:

Given that

$$\begin{gathered} \frac{T_{10}}{T_{11}}=\frac{45}{22} \\ \frac{T_{9+1}}{T_{10+1}}=\frac{45}{22} \\ \text{We know that for}{(a+b)}^n,n\text{th term is}{T}_{r+1}={}^n{C}_r{a}^{n-r}{b}^r \\ \Rightarrow \frac{{}^{20}{C}_9{\left(2\right)}^{20-9}{\left(-3x^3\right)}^9}{{}^{20}{C}_{10}{\left(2\right)}^{20-10}{\left(-3x^3\right)}^{10}}=\frac{45}{22} \\ \Rightarrow \frac{{}^{20}{C}_9{\left(2\right)}^{20-9}{\left(-3x^3\right)}^9}{{}^{20}{C}_{10}{\left(2\right)}^{20-10}{\left(-3x^3\right)}^{10}}=\frac{45}{22} \\ \Rightarrow \frac{{\displaystyle \frac{20!}{9!11!}}{\left(2\right)}^{11}{\left(-3\right)}^9{x}^{27}}{{\displaystyle \frac{20!}{10!10!}}{\left(2\right)}^{10}{\left(-3\right)}^{10}{x}^{30}}=\frac{45}{22}\left[\because {}^{n}C_r=\frac{n!}{r!(n-r)!}\right] \\ \Rightarrow \frac{10}{11}\frac{2}{\left(-3x^3\right)}=\frac{45}{22} \\ \Rightarrow 3x^3=-\frac{8}{9} \\ \Rightarrow x^3=-\frac{8}{27} \\ \Rightarrow x=-\frac{2}{3}[\text{taking cube root both sides]} \end{gathered}$$

Hence, option A is correct.