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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the middle term in the expansion of $(\large\frac{x}{a}-\frac{a}{x})^{10}$

$\begin{array}{1 1}(A)\;251\\(B)\;252\\(C)\;-252\\(D)\;254\end{array} $

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1 Answer

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Toolbox:
  • If $n+1$ is odd,the middle term is $\large\frac{(n+1)+1}{2}$$+h=\large\frac{n+2}{2}$$+h$
  • $T_{r+1}=(-1)^rnC_ra^{n-r}b^r$
Number of terms in the expansion is 10+1=11
Hence the middle term $\large\frac{10+2}{2}$$=6^{th}$
$T_{r+1}=(-1)^{r} C(10,r) (\large\frac{x}{a})^{10-r}(\large\frac{a}{x})^r$
$T_6=(-1)^5C(10,5)(\large\frac{x^5}{a^5}).\frac{a^5}{x^5}$
$\Rightarrow (-1).\large\frac{10!}{5!5!}$
$\Rightarrow (-1).\large\frac{10\times 9\times 8\times \times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 5!}$
$\Rightarrow -252$
Hence (C) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

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