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 :

$\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r$ <br/><br/>$$ \frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6} $$ <br/><br/>$\left[\because r\right.$th term from end in the expansion of $(x+y)^n=r$th term from beginning in the expansion of $\left.(y+x)^n\right]$ <br/><br/>$$ \Rightarrow \frac{{ }^n C_4(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)^{4 / 4}}{{ }^n C_4\left(\frac{1}{3}\right)^{\frac{n-4}{4}}(2)^{4 / 4}}=\frac{\sqrt{6}}{1} $$ <br/><br/>$$ \begin{aligned} &\Rightarrow (2)^{\frac{n-8}{4}}(3)^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow 6^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow \frac{n-8}{4} =\frac{1}{2} \\\\ &\Rightarrow n-8=2 \Rightarrow n =10 \end{aligned} $$ <br/><br/>$$ \therefore T_3={ }^{10} C_2(\sqrt[4]{2})^8\left(\frac{1}{\sqrt[4]{3}}\right)^2=45(2)^{\frac{8}{4}} \frac{1}{3^{1 / 2}} $$ <br/><br/>$=45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3}$