Verified Solution • 2025–26 Session moderate

QJEEM23Ma4646: JEE Main 2023 Mathematics - Class 11 JEEM23Ma4646 Solution: If the ratio of the fifth term from the beginning to the fifth…

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (24+134)n\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}} is 6:1\sqrt{6}: 1 , then the third term from the beginning is :

Multiple Choice Options

  • A 30230 \sqrt{2}
  • B 60360 \sqrt{3}
  • C 60260 \sqrt{2}
  • D 30330 \sqrt{3}

Step-by-Step Verified Solution

Tr+1=nCrxn−rar\mathrm{T}_{r+1}={ }^n \mathrm{C}_r x^{n-r} a^r

T5T5′=nC4×((2)14)n−4(131/4)4nC4(131/4)n−4(21/4)4=6\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}

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

⇒nC4(2)n−44(13)4/4nC4(13)n−44(2)4/4=61\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}

amp;⇒(2)n−84(3)n−84=61/2amp;⇒6n−84=61/2amp;⇒n−84=12amp;⇒n−8=2⇒n=10\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}

∴T3=10C2(24)8(134)2=45(2)84131/2\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)×33=603=45(4) \times \frac{\sqrt{3}}{3}=60 \sqrt{3}

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