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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The position of the term independent of $x$ in the expansion of $\big(\sqrt{\large\frac{x}{3}}+\large\frac{3}{2x^2}\big)^{10}$ is

$\begin{array}{1 1}(A)\;10C_1\\(B)\;\large\frac{5}{12}\\(C)\;1\\(D)\;\text{Third term}\end{array} $

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1 Answer

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  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_{r+1}=10C_r \big(\sqrt{\large\frac{x}{3}}\big)^{10-r}.(\large\frac{3}{2x^2})^r$
$\Rightarrow 10C_r \big((\large\frac{x}{3})^{1/2}\big)^{10-r}.\big(\large\frac{3}{2x^2}\big)^r$
$\Rightarrow 10C_r \big[\large\frac{x}{3}^{\Large\frac{10-r}{2}}\big(\large\frac{3}{2^r x^{2r}}\big)$
$\Rightarrow 10C_r (\large\frac{1}{\sqrt 3})^{10-r}(\large\frac{3}{2})^r.$$x^{5-\Large\frac{r}{2}-\normalsize 2r}$
Let $T_{r+1}$ be the term independent of $x$
$5-\large\frac{r}{2}$$-2r=0$
$\large\frac{10-r-4r}{2}=$$0$
$10-5r=0$
$r=2$
Hence the third term is independent of $x$
Hence (D) is the correct answer.
answered Jun 25, 2014 by sreemathi.v
 

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