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  • Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is root 6 : 1

Q8.    Find n, 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 )^nis \sqrt6 :1

Answers (1)

best_answer

Given, the expression 

\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}} \right )^n

Fifth term from the beginning  is 

T_5=^nC_4(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^4

T_5=^nC_4\frac{(\sqrt[4]{2})^n}{(\sqrt[4]{2})^4}\times\frac{1}{3}

T_5=\frac{n!}{4!(n-4)!}\times\frac{(\sqrt[4]{2})^n}{2}\times\frac{1}{3}

And Fifth term from the end is,

T_{n-5}=^nC_{n-4}(\sqrt[4]{2})^4\left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}

T_{n-5}=^nC_{n-4}(\sqrt[4]{2})^4\left ( \frac{(\sqrt[4]{3})^4}{(\sqrt[4]{3})^n} \right )

T_{n-5}=\frac{n!}{4!(n-4)!}\times2\times\left ( \frac{3}{(\sqrt[4]{3})^n} \right )

Now, As given in the question,

T_5:T_{n-5}=\sqrt{6}:1

So,

\left(\frac{n!}{4!(n-4)!}\times\frac{(\sqrt[4]{2})^n}{2}\times\frac{1}{3}\right):\left( \frac{n!}{4!(n-4)!}\times2\times\left ( \frac{3}{(\sqrt[4]{3})^n} \right )\right)=\sqrt{6}:1

From Here ,

\frac{(\sqrt[4]{2})^n}{6}:\frac{6}{(\sqrt[4]{3})^n}=\sqrt{6}:1

\frac{(\sqrt[4]{2})^n(\sqrt[4]{3})^n}{6\times6}=\sqrt{6}

(\sqrt[4]{6})^n=36\sqrt{6}

6^{\frac{n}{4}}=6^{\frac{5}{2}}

From here,

\frac{n}{4}=\frac{5}{2}

n=10

Hence the value of n is 10.

 

 

 

Posted by

Pankaj Sanodiya

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