If the ratio of the fifth term from the beginning to the fifth term from the end in the ex…

Question

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $$\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$$ is $$\sqrt{6}: 1$$, then the third term from the beginning is :

Accepted Answer

$$
\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r
$$

$$
\frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6}
$$

$\left[\because r\right.$th term from end in the expansion of $(x+y)^n=r$th term from beginning in the expansion of $\left.(y+x)^n\right]$

$$
\Rightarrow \frac{{ }^n C_4(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)^{4 / 4}}{{ }^n C_4\left(\frac{1}{3}\right)^{\frac{n-4}{4}}(2)^{4 / 4}}=\frac{\sqrt{6}}{1}
$$

$$
\begin{aligned}
&\Rightarrow  (2)^{\frac{n-8}{4}}(3)^{\frac{n-8}{4}} =6^{1 / 2} \\
&\Rightarrow  6^{\frac{n-8}{4}}  =6^{1 / 2} \\
&\Rightarrow  \frac{n-8}{4}  =\frac{1}{2} \\
&\Rightarrow  n-8=2 \Rightarrow n  =10
\end{aligned}
$$

$$
\therefore T_3={ }^{10} C_2(\sqrt[4]{2})^8\left(\frac{1}{\sqrt[4]{3}}\right)^2=45(2)^{\frac{8}{4}} \frac{1}{3^{1 / 2}}
$$

$$
=45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3}
$$

Solutions · 1