Question

Let $P$ be the $7^{\text {th }}$ term from the beginning and $Q$ be the $7^{\text {th }}$ term from the end in the expansion of $\left(\sqrt[3]{3}+\frac{1}{\sqrt[3]{4}}\right)^{n},$ where $n \in N$. If $\frac{Q}{P}=12,$ then $n$ is equal to ____

Answer 1

$P={ }^{n} C_{6} \cdot\left(3^{\frac{1}{3}}\right)^{n-6} \cdot\left(4^{-\frac{1}{3}}\right)^{6}$
$Q={ }^{n} C_{n-6} \cdot\left(3^{\frac{1}{3}}\right)^{6} \cdot\left(4^{-\frac{1}{3}}\right)^{n-6}$
$\therefore \frac{Q}{P}=12 P \Rightarrow (12)^{\frac{n-6}{3}}=(12)^{1}$
$\Rightarrow \frac{n-6}{3}=1 \Rightarrow n=9$