# If in the expansion of $(x+a)^{15},$ the $11^{th}$ term is $G.M.$ of $8^{th}$ & $12^{th}$ terms, then which term is the greatest term?

$\begin{array}{1 1} T_6 \\ T_7 \\ T_8 \\ T_9 \end{array}$

## 1 Answer

Given: $(T_{11})^2=T_8.T_{12}$ in the expansion of $(x+a)^{15}$
$\Rightarrow\:\big[^{15}C_{10}.x^5.a^{10}\big]^2=(^{15}C_7.x^8.a^7)\times(^{15}C_{11}.x^4.a^{11})$
$\Rightarrow\:(^{15}C_{10})^2.a^2=^{15}C_{7}.^{15}C_{11}.x^2$
$\Rightarrow\:\large\frac{a^2}{75}=\frac{x^2}{77}$
$\Rightarrow\:\large\frac{a}{x}=\sqrt {\frac{75}{77}}$
Let the $(r+1)^{th}$ term be the highest term.
Then $\large\frac{T_{r+1}}{T_r}\geq 1$

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