If the eleventh term in the binomial expansion of $(x+a)^{15}$ is the geometric mean of the eighth and…

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
  1. $7^{\text {th }}$ term
  2. $8^{\text {th }}$ term
  3. $9^{\text {th }}$ term
  4. $10^{\text {th }}$ term

Solution

From the given condition $T_{11}=\sqrt{T_8 T_{12}}$ $\begin{aligned} & \Rightarrow\left({ }^{15} C_{10} x^5 a^{10}\right)=\left({ }^{15} C_7 x^8 a^7\right)\left({ }^{15} C_{11} x^4 a^{11}\right) \\ & \Rightarrow \frac{x}{a}=\sqrt{\frac{77}{75}} \end{aligned}$
Now, greatest term $= \begin{cases}T_p \text { and } T_{p+1} ; & \text { if } \frac{n+1}{\left|\frac{x}{a}\right|+1}=p \text { is an integer } \\ T_{q+1} & ; \text { if } \frac{n+1}{\left|\frac{x}{a}\right|+1} \text { is non integer and } \in(q, q+1)\end{cases}$
So, $\frac{n+1}{\left|\frac{x}{a}\right|+1}=\frac{16}{\sqrt{\frac{77}{75}}+1} \in(7,8)$ $\therefore$ Greatest term $=\mathrm{T}_8$ i., e., $8^{\text {th }}$ term.

Asked in: AP EAMCET 2024 (21 May Shift 1)

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