Question

In $(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}}{)}^n$ if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}$, then n =

• A. 7
• B. 8
• C. 9
• D. None of these

Answer 1

The correct option is C

9

$\frac{1}{6}=\frac{{}^n{C}_6{\left(2^{\frac{1}{3}}\right)}^{n-6}{\left(3^{-\frac{1}{3}}\right)}^6}{{}^n{C}_{n-6}{\left(2^{\frac{1}{3}}\right)}^6{\left(3^{-\frac{1}{3}}\right)}^{n-6}}$

$6^{-1}=6^{\frac{1}{3}(n-12)}$

$n-12=-3$

$n=9$