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Find $x$ if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(4\sqrt 2+\large\frac{1}{4\sqrt 3})^n$ is $\sqrt 6 :1$

$\begin{array}{1 1}(A)\;8\\(B)\;9\\(C)\;10\\(D)\;11\end{array} $

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  • $T_{r+1}$=General term =$nC_ra^{n-r}b^r$
  • $\therefore (a+b)^n=\sum\limits_{r=0}^n nC_ra^{n-r}b^r$
$T_5$ in $\big[2^{\Large\frac{1}{4}}+\large\frac{1}{3^{\Large\frac{1}{4}}}\big]^n$$=nC_4(2^{1/4})^{n-4}.\big(\large\frac{1}{3^{1/4}}\big)^4$
$\Rightarrow nC_42^{\Large\frac{n-4}{4}}.\large\frac{1}{3}$-------(1)
Total number of terms =n+1
Fifth term from the end =$[(n+1)-5+1)]^{th}$ term from the beginning.
$\Rightarrow (n-3)^{th}$ term from the beginning.
$\Rightarrow nC_{n-4}(2^{1/4})^{n-(n-4)}\big(\large\frac{1}{3^{1/4}}\big)^{n-4}$
$\Rightarrow nC_4.2(\large\frac{1}{3})^{\Large\frac{n-4}{4}}$------(2)
Dividing (1) by (2)
$\large\frac{nC_42^{\Large\frac{n-4}{4}}.\large\frac{1}{3}}{nC_4.2(\large\frac{1}{3})^{\Large\frac{n-4}{4}}}=\frac{\sqrt 6}{1}$
Or $\large\frac{2^{\Large\frac{n}{4}-2}}{(1/3)^{\Large\frac{n}{4}-2}}=\frac{\sqrt 6}{1}$
Or $2^{\Large\frac{n}{4}-2}.3^{\Large\frac{n}{4}-2}=6^{\Large\frac{1}{2}}$
$\therefore \large\frac{n}{4}$$-2=\large\frac{1}{2}$
$\therefore \large\frac{n}{4}=\frac{5}{2}$
$n=\large\frac{5}{2}$$\times 4$
Hence (C) is the correct answer.
answered Jun 24, 2014 by sreemathi.v

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