In the expansion of $(\mathrm{x}+\mathrm{y}{)}^{15}$ the eleventh term is geometric mean of ninth and twelfth terms then k^th term of the expansion must be greatest then the value of k is

In the expansion of $(\mathrm{x}+\mathrm{y}{)}^{15}$ the eleventh term is geometric mean of ninth and twelfth terms then k^th term of the expansion must be greatest then the value of k is

$\begin{array}{l}{\mathrm{T}}_{11}{=}^{15}{\mathrm{C}}_{10}{\mathrm{x}}^5{\mathrm{y}}^{10}, {\mathrm{T}}_9{=}^{15}{\mathrm{C}}_8{\mathrm{x}}^7{\mathrm{y}}^8,{\mathrm{T}}_{12}{=}^{15}{\mathrm{C}}_{11}{\mathrm{x}}^4{\mathrm{y}}^{11}\\ \Rightarrow {\mathrm{T}}_{11}=\sqrt{{\mathrm{T}}_9{\mathrm{T}}_{12}}\\ { }^{15}{\mathrm{C}}_{10}{\mathrm{x}}^5{\mathrm{y}}^{10}=\sqrt{{ }^{15}{\mathrm{C}}_8{\times }^{15}{\mathrm{C}}_4{\mathrm{x}}^{11}{\mathrm{y}}^{19}}\Rightarrow \mathrm{y}{\left({ }^{15}{\mathrm{C}}_{10}\right)}^2{=}^{15}{\mathrm{C}}_7{\times }^{15}{\mathrm{C}}_4\left(\mathrm{x}\right)\\ \Rightarrow \frac{7!8!}{5!10!}\times \frac{11}{5}=\frac{\mathrm{x}}{\mathrm{y}}\\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{75}{77}\\ \\ \Rightarrow \mathrm{r}=\left[\frac{(\mathrm{n}+1)\left|\frac{\mathrm{y}}{\mathrm{x}}\right|}{\left|\frac{\mathrm{y}}{\mathrm{x}}\right|+1}\right]=\left[\frac{16\times \frac{75}{77}}{\frac{75}{77}+1}\right]=7\end{array}$

${\mathrm{T}}_8\text{ is Numirically greatest term}$