Question

Find the value of x, if the ratio of 10th term to 11th term of the expansion $(2-3x^3{)}^{20}$ is 45 : 22.

Or

Find the value of a, so that the term independent of x in ${(\sqrt{x}+\frac{a}{x^2})}^{10}$ is 405.

Answer 1

We have, the ratio of 10th term to 11th term of the expansion $(2-3x^3{)}^{20}$ is 45 : 22.

$\therefore \frac{T_{10}}{T_{11}}=\frac{45}{22}$

Now, $T_{10}={}^{20}{C}_9\left(2{)}^{20-9}\right(-3x^3{)}^9$

$\therefore T_{10}={}^{20}{C}_9{2}^{11}(-3{)}^9(x^3{)}^9$

$\therefore T_{10}=(-1{)}^9{}^{20}{C}_9{2}^{11}{.3}^9.x^{27}$

and $T_{11}={}^{20}{C}_{10}\left(2{)}^{20-10}\right(-3x^3{)}^{10}$

$\therefore T_{11}={}^{20}{C}_{10}{2}^{10}{.3}^{10}.x^{30}$

We have given, $\frac{T_{10}}{T_{11}}=\frac{45}{22}$

$\therefore \frac{(-1{)}^9{}^{20}{C}_9{.2}^{11}{.3}^9.x^{27}}{{11.3}^{10}.x^{30}}=\frac{45}{22}$

$\Rightarrow \frac{(-1)\times 10\times 2}{(20+1-10)\times 3\times x^3}=\frac{45}{22}[\because \frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{r}{n+1-r}]$

$\Rightarrow x^3=\frac{(-1)\times 10\times 2\times 22}{11\times 3\times 45}=\frac{-8}{27}$

$\Rightarrow x=-\frac{2}{3}$

Or

Let $(r+1)$ term in the expansion of ${(\sqrt{x}+\frac{a}{x^2})}^{10}$ be independent of x.

Now, $T_{r+1}={}^{10}{C}_r(\sqrt{x}{)}^{10-r}{\left(\frac{a}{x^2}\right)}^r$

$\Rightarrow T_{r+1}={}^{10}{C}_r\left(x{)}^{5-\frac{r}{2}}\right(a{)}^r(X{)}^{-2r}$

$\Rightarrow T_{r+1}={}^{10}{C}_r(X{)}^{5-\frac{r}{2}-2r}{a}^r...(i)$

This will be independent of x, if

$5-\frac{r}{2}-2r=0$

$\Rightarrow 5r=10\Rightarrow r=2$

On substituting the value of r in Eq. (i), we get

$T_3={}^{10}{C}_2{a}^2=45a^2$

It is given that, the term independent of x is 405.

$\therefore 45a^2=405\Rightarrow a^2=9$

$\Rightarrow a=\pm 3$