Question

The fourth term in the expansion of ${(px+\frac{1}{x})}^n$ is $\frac{5}{2}$. Then,

• A. $n=6$
• B. $n=7$
• C. $p=\frac{1}{2}$
• D. $p=\frac{1}{4}$

Answer 1

Answer: (A), (C)

$n=6$
$p=\frac{1}{2}$

In the above question
$T_{r+1}={}^n{C}_r{x}^{n-2r}{p}^{n-r}$
$T_{3+1}={}^n{C}_3{x}^{n-6}{p}^{n-3}$
$=\frac{5}{2}$
Since it is independent of x,
$n-6=0$
$n=6$
Hence, ${}^6{C}_3{p}^3=\frac{5}{2}$
$20p^3=\frac{5}{2}$
$p^3=\frac{1}{8}$
$p=\frac{1}{2}$