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In the binomial expansion of $(a-b)^n,\:n\geq 5$ the sum of $5^{th}$ and $6^{th}$ term is $0$, then $\large\frac{a}{b}$=?

$\begin{array}{1 1} \frac{6}{n-5} \\ \frac{n-5}{6} \\ \frac{5}{n-4} \\ \frac{n-4}{5} \end{array}$

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1 Answer

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Toolbox:
  • $T_{r+1}=^nC_r x^{n-r}.y^r$ in the expansion $(x+y)^n$
$T_5=^nC_4 a^{n-4}b^4$
$T_6=-^nC_5 a^{n-5}b^5$
Given $T_5+T_6=0$
$\Rightarrow\:^nC_4 a^{n-4}b^4-^nC_5 a^{n-5}b^5=0$
$\Rightarrow\:^nC_4 a^{n-4}b^4=^nC_5 a^{n-5}b^5$
$\Rightarrow\:\large\frac{1}{n-4}=$$\large\frac{b}{5a}$
$\Rightarrow\:\large\frac{b}{a}=\frac{5}{n-4}$
answered Sep 16, 2013 by rvidyagovindarajan_1
 

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