Question

If the ratio of 7^th term from the beginning to the seventh term from the end in the expansion of ($\sqrt[3]{32}+\frac{1}{\sqrt{3}}{)}^n$is $\frac{1}{6}$, then n is

• A. 9
• B. 6
• C. 12
• D. None of these

Answer 1

The correct option is A

9

$T_7$ from end of $(a+b{)}^n$ is same as $T_7$ from \space beginning \space of $(b+a{)}^n$

$\frac{T_{7}frombeginning}{T_{7}fromend}$ of $(a+b{)}^n$

or in $(2^{\frac{1}{3}}+\frac{1}{3^{\frac{1}{3}}}{)}^n$ or $\frac{T^{7}of(a+b{)}^n}{T_{7}of(b+a{)}^n}$ = $\frac{1}{6}$

or $(2^{\frac{1}{3}}.3^{\frac{1}{3}}{)}^{n-12}$ = 6^{-1} ∴ $\frac{n-12}{3}$ = -1 ∴ n = 9