First slide

Binomial theorem for positive integral Index

Question

If the 4th term in the expansion of $(ax + 1 / x)^{n} is \frac{5}{2}$ , then

Moderate

A

$a = \frac{1}{2}$
B

n=8
C

$a = \frac{2}{3}$
D

n=6

Solution

It is given that the fourth term in the expansion of
$\begin{array}{l} {(ax + \frac{1}{x})}^{n} is \frac{5}{2}; therefore \\ ^{n} C_{3} (ax)^{n - 3} {(\frac{1}{x})}^{3} = \frac{5}{2} \Rightarrow^{n} C_{3} a^{n - 3} x^{n - 6} = \frac{5}{2} \end{array}$
[ R.H.S. is independent of x]
Putting n=6in(1),we get $^{6} C_{3} a^{3} = \frac{5}{2} or a^{3} = \frac{1}{8} or a = \frac{1}{2}$

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