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Find n, if the ratio of the fifth term from the beginning to t

Question

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(42 + \frac{1}{4 3})^{n}$ is $6 : 1$

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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

(42 + \frac{1}{4 3})^{n}

6 : 1

6 mins ago

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6 mins ago

Text solutionVerified

Solution:

r + 1^{t h}

term in the given expression can be given as,

T_{r + 1} = C (n, r) (42)^{n - r} (\frac{1}{4 3})^{r}

∴ T_{5} = C (n, 4) (42)^{n - 4} (\frac{1}{4 3})^{4}

Fifth term from end will be

T_{n - 3}

∴ T_{n - 3} = C (n, n - 4) (42)^{4} (\frac{1}{4 3})^{n - 4}

Now, we are given,

\frac{T _{5}}{T _{n - 3}} = \frac{6}{1}

∴ \frac{C ( n , 4 ) ( 4 2 ) ^{n - 4} ( \frac{1}{4 3} ) ^{4}}{C ( n , n - 4 ) ( 4 2 ) ^{4} ( \frac{1}{4 3} ) ^{n - 4}} = \frac{6}{1}

We know,

C (n, 4) = C (n, (n - 4))

∴ \frac{2 ^{\frac{n}{4} - 1} 3 ^{- 1}}{2 ^{1} 3 ^{1 - \frac{n}{4}}} = \frac{6}{1}

\Rightarrow 2^{\frac{n}{4} - 2} 3^{\frac{n}{4} - 2} = 2^{\frac{1}{2}} 3^{\frac{1}{2}}

∴ \frac{n}{4} - 2 = \frac{1}{2}

\Rightarrow \frac{n}{4} = \frac{5}{2} \Rightarrow n = 10

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Question Text	Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(42 + \frac{1}{4 3})^{n}$ is $6 : 1$
Answer Type	Text solution:1
Upvotes	150