If fourth term in the expansion (ax+x1)n is 25 then find the values of a and n

(ax+x1)n=nC0(ax)n+nC1(ax)n−1(x1)+nC2(ax)n−2(x1)2+nC3(ax)n−3(x1)3+...Fourth term=t4=nC3(ax)n−3(x1)3⇒t4=nC3(ax)n−3(x31)⇒nC3an−3xn−6=25x0Comparing the coefficeint of x⇒n−6=0⇒n=6So,C3nan−3=25⇒C36a6−3=25⇒C36a3=25⇒3!3!6!a3=25⇒3!×3×26×5×4×3!a3=25⇒a3=81⇒a=21∴n=6 and a=21

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If fourth term in the expansion {(ax+{1}{x})}^{n} is {5}{2} th

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If fourth term in the expansion $(a x + \frac{1}{x})^{n}$ is $\frac{5}{2}$ then find the values of $a$ and $n$

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Michael Discussed

If fourth term in the expansion

(a x + \frac{1}{x})^{n}

\frac{5}{2}

then find the values of

a

and

n

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

(a x + \frac{1}{x})^{n}

=^{n} C_{0} (a x)^{n} +^{n} C_{1} (a x)^{n - 1} (\frac{1}{x}) +^{n} C_{2} (a x)^{n - 2} (\frac{1}{x})^{2} +^{n} C_{3} (a x)^{n - 3} (\frac{1}{x})^{3} + ...

Fourth term

= t_{4} =^{n} C_{3} (a x)^{n - 3} (\frac{1}{x})^{3}

\Rightarrow t_{4} =^{n} C_{3} (a x)^{n - 3} (\frac{1}{x ^{3}})

\Rightarrow^{n} C_{3} a^{n - 3} x^{n - 6} = \frac{5}{2} x^{0}

Comparing the coefficeint of

x

\Rightarrow n - 6 = 0

\Rightarrow n = 6

So,

_{C_{3}}^{n} a^{n - 3} = \frac{5}{2}

\Rightarrow_{C_{3}}^{6} a^{6 - 3} = \frac{5}{2}

\Rightarrow_{C_{3}}^{6} a^{3} = \frac{5}{2}

\Rightarrow \frac{6 !}{3 ! 3 !} a^{3} = \frac{5}{2}

\Rightarrow \frac{6 \times 5 \times 4 \times 3 !}{3 ! \times 3 \times 2} a^{3} = \frac{5}{2}

\Rightarrow a^{3} = \frac{1}{8}

\Rightarrow a = \frac{1}{2}

∴ n = 6

and

a = \frac{1}{2}

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Question Text	If fourth term in the expansion $(a x + \frac{1}{x})^{n}$ is $\frac{5}{2}$ then find the values of $a$ and $n$
Answer Type	Text solution:1
Upvotes	150

If fourth term in the expansion (ax+x1​)n is 25​ then find the values of a and n

Text solutionVerified

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If fourth term in the expansion $(a x + \frac{1}{x})^{n}$ is $\frac{5}{2}$ then find the values of $a$ and $n$