JEE MAIN - Mathematics (2023 - 6th April Morning Shift - No. 12)
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 :
$$30 \sqrt{2}$$
$$60 \sqrt{3}$$
$$60 \sqrt{2}$$
$$30 \sqrt{3}$$
Explanation
$$
\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r
$$
$$ \frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6} $$
$\left[\because r\right.$th term from end in the expansion of $(x+y)^n=r$th term from beginning in the expansion of $\left.(y+x)^n\right]$
$$ \Rightarrow \frac{{ }^n C_4(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)^{4 / 4}}{{ }^n C_4\left(\frac{1}{3}\right)^{\frac{n-4}{4}}(2)^{4 / 4}}=\frac{\sqrt{6}}{1} $$
$$ \begin{aligned} &\Rightarrow (2)^{\frac{n-8}{4}}(3)^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow 6^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow \frac{n-8}{4} =\frac{1}{2} \\\\ &\Rightarrow n-8=2 \Rightarrow n =10 \end{aligned} $$
$$ \therefore T_3={ }^{10} C_2(\sqrt[4]{2})^8\left(\frac{1}{\sqrt[4]{3}}\right)^2=45(2)^{\frac{8}{4}} \frac{1}{3^{1 / 2}} $$
$$ =45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3} $$
$$ \frac{T_5}{T_5^{\prime}}=\frac{{ }^n C_4 \times\left((2)^{\frac{1}{4}}\right)^{n-4}\left(\frac{1}{3^{1 / 4}}\right)^4}{{ }^n C_4\left(\frac{1}{3^{1 / 4}}\right)^{n-4}\left(2^{1 / 4}\right)^4}=\sqrt{6} $$
$\left[\because r\right.$th term from end in the expansion of $(x+y)^n=r$th term from beginning in the expansion of $\left.(y+x)^n\right]$
$$ \Rightarrow \frac{{ }^n C_4(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)^{4 / 4}}{{ }^n C_4\left(\frac{1}{3}\right)^{\frac{n-4}{4}}(2)^{4 / 4}}=\frac{\sqrt{6}}{1} $$
$$ \begin{aligned} &\Rightarrow (2)^{\frac{n-8}{4}}(3)^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow 6^{\frac{n-8}{4}} =6^{1 / 2} \\\\ &\Rightarrow \frac{n-8}{4} =\frac{1}{2} \\\\ &\Rightarrow n-8=2 \Rightarrow n =10 \end{aligned} $$
$$ \therefore T_3={ }^{10} C_2(\sqrt[4]{2})^8\left(\frac{1}{\sqrt[4]{3}}\right)^2=45(2)^{\frac{8}{4}} \frac{1}{3^{1 / 2}} $$
$$ =45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3} $$
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