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If the 9 th and 10th terms are the numerically greatest terms in the expansion of $(5 x-6 y)^n$ when $x=2 / 5$ and $y=1 / 2$, then the absolute value of the middle terms of that expansion is
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$14 C_7 6^7$
Given, 9th and 10th terms are numerically greatest terms in the expansion of $(5 x-6 y)^n$
$9 \leq \frac{(n+1)\left(\frac{6 y}{5 x}\right)}{1+\left(\frac{6 y}{5 x}\right)}$
Here, $y=\frac{1}{2}$ and $x=\frac{2}{5}$
$\because \quad \frac{y}{x}=\frac{5}{4} \Rightarrow \frac{6 y}{5 x}=\frac{3}{2}$
$9 \leq \frac{(n+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}$
$9 \leq \frac{3(n+1)}{5}$
$n+1 \geq 15$
$n \geq 14$
$\because \quad n=14$
Middle term of $(5 x-6 y)^{14}$
$=\left|{ }^{14} C_7(5 x)^7(-6 y)^7\right|$
$=\left|{ }^{14} C_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} C_7 \times 6^7$
$9 \leq \frac{(n+1)\left(\frac{6 y}{5 x}\right)}{1+\left(\frac{6 y}{5 x}\right)}$
Here, $y=\frac{1}{2}$ and $x=\frac{2}{5}$
$\because \quad \frac{y}{x}=\frac{5}{4} \Rightarrow \frac{6 y}{5 x}=\frac{3}{2}$
$9 \leq \frac{(n+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}$
$9 \leq \frac{3(n+1)}{5}$
$n+1 \geq 15$
$n \geq 14$
$\because \quad n=14$
Middle term of $(5 x-6 y)^{14}$
$=\left|{ }^{14} C_7(5 x)^7(-6 y)^7\right|$
$=\left|{ }^{14} C_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} C_7 \times 6^7$
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