If the eleventh term in the binomial expansio | vedclass

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(C) The general term of $(x+a)^{15}$ is $T_{r+1} = {}^{15}C_r x^{15-r} a^r$.
Given $T_{11} = \sqrt{T_8 T_{12}}$.
Substituting the terms: ${}^{15}C_{

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$9^{	ext{th}}$ term

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(C) The general term of $(x+a)^{15}$ is $T_{r+1} = {}^{15}C_r x^{15-r} a^r$.
Given $T_{11} = \sqrt{T_8 T_{12}}$.
Substituting the terms: ${}^{15}C_{10} x^5 a^{10} = \sqrt{({}^{15}C_7 x^8 a^7)({}^{15}C_{11} x^4 a^{11})}$.
${}^{15}C_{10} x^5 a^{10} = \sqrt{{}^{15}C_7 \cdot {}^{15}C_{11} x^{12} a^{18}}$.
${}^{15}C_{10} x^5 a^{10} = \sqrt{6435 \cdot 1365} x^6 a^9$.
$3003 x^5 a^{10} = 2964.7 x^6 a^9 \Rightarrow \frac{x}{a} = \frac{3003}{2964.7} \approx 1.013$.
The greatest term $T_{r+1}$ occurs when $\frac{r}{n-r+1} \le \frac{x}{a} \le \frac{n-r+1}{r}$ is not applicable here,we use $r = \lfloor \frac{(n+1)|x|}{|x|+|a|} \rfloor$.
$r = \lfloor \frac{16 \cdot 1.013}{1.013+1} \rfloor = \lfloor \frac{16.208}{2.013} \rfloor = \lfloor 8.05 \rfloor = 8$.
Thus,$T_{8+1} = T_9$ is the greatest term.

If the eleventh term in the binomial expansion of $(x+a)^{15}$ is the geometric mean of the eighth and twelfth terms,then the greatest term in the expansion is

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