Logo
Jeeify
MCQMediumJEE 2023Simple Applications

JEE Mathematics 2023 Question with Solution

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (2+13)n\left(\sqrt{2}+\frac{1}{\sqrt{3}}\right)^n is 6:1\sqrt{6}:1, then the third term from the beginning is :

  • A

    30230\sqrt{2}

  • B

    60260\sqrt{2}

  • C

    30330\sqrt{3}

  • D

    60360\sqrt{3}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (2+13)n\left(\sqrt{2}+\frac{1}{\sqrt{3}}\right)^n is 6:1\sqrt{6}:1.

Find: The third term from the beginning.

Using the expressions shown in the solution,

nC4(214)n−4(3−14)4nC4(2−14)n−4(314)4=6\frac{{}^nC_4\left(2^{\frac{1}{4}}\right)^{n-4}\left(3^{-\frac{1}{4}}\right)^4}{{}^nC_4\left(2^{-\frac{1}{4}}\right)^{n-4}\left(3^{\frac{1}{4}}\right)^4}=\sqrt{6}

So,

(2143−14)n−8=6\left(\frac{2^{\frac{1}{4}}}{3^{-\frac{1}{4}}}\right)^{n-8}=\sqrt{6}

Hence,

6n−84=66^{\frac{n-8}{4}}=\sqrt{6}

Equating powers of 66,

n−84=12\frac{n-8}{4}=\frac{1}{2}

Therefore,

n−8=2n-8=2

so

n=10n=10

Now the third term from the beginning is

T3=10C2(214)8(3−14)2T_3={} ^{10}C_2\left(2^{\frac{1}{4}}\right)^8\left(3^{-\frac{1}{4}}\right)^2

Simplifying,

T3=10C2×(2)4×13T_3={} ^{10}C_2\times (\sqrt{2})^4\times \frac{1}{\sqrt{3}}

From the extracted solution,

T3=603T_3=60\sqrt{3}

Therefore, the third term from the beginning is 60360\sqrt{3}. Hence, the correct option is D.

Term Identification

For a binomial expansion (a+b)n(a+b)^n, the fifth term from the beginning involves r=4r=4, and the fifth term from the end corresponds to the symmetric term obtained from the other side. The given ratio leads to the power comparison shown in the solution.

After simplifying the ratio, we get

6n−84=6=6126^{\frac{n-8}{4}}=\sqrt{6}=6^{\frac{1}{2}}

which gives

n=10 $$.

Then the third term from the beginning is obtained by taking r=2r=2:

T3=10C2(2)8(13)2T_3={} ^{10}C_2\left(\sqrt{2}\right)^8\left(\frac{1}{\sqrt{3}}\right)^2

which simplifies to the required value 60360\sqrt{3} as stated in the solution.

Common mistakes

  • Using the fifth term from the end incorrectly. In a binomial expansion, terms from the end must be mapped carefully using symmetry. Instead of guessing its form, write the general term and then identify the correct corresponding index.

  • Cancelling the combinatorial factors and powers incorrectly in the ratio. The coefficients nC4{}^nC_4 cancel here, but the exponents of 22 and 33 must still be handled carefully using laws of indices.

  • Finding the third term wrongly as 10C3{}^{10}C_3 instead of 10C2{}^{10}C_2. The third term from the beginning corresponds to r=2r=2 in the general term Tr+1T_{r+1}, not r=3r=3.

Practice more Simple Applications questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trialMore on Simple Applications

Related questions