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In the binomial expansion of $(a-b)^n, n \geq 5$, the sum of $5^{\text {th }}$ and $6^{\text {th }}$ terms is zero, then $\frac{a}{b}$ equals
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The correct answer is:
$\frac{n-4}{5}$
$\frac{n-4}{5}$
${ }^n C_4 a^{n-4}(-b)^4+{ }^n C_5 a^{n-5}(-b)^5=0$
$\Rightarrow\left(\frac{a}{b}\right)=\frac{n-5+1}{5}$.
$\Rightarrow\left(\frac{a}{b}\right)=\frac{n-5+1}{5}$.
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