Question

If the fourth term in the expansion of ${(px+\frac{1}{x})}^n$ is $\frac{5}{2}$, then $(n,p)=$

• A. $(3,\frac{1}{2})$
• B. $(6,\frac{1}{2})$
• C. $(5,\frac{1}{2})$
• D. $(6,2)$

Answer 1

Answer: (B)

$(6,\frac{1}{2})$

Given expression is $(px+x^{-1}){}^n$
General term $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Applying to the above question, we get

$T_{3+1}={}^n{C}_3(px{)}^{n-3}{x}^{-3}$
$={}^n{C}_3(p{)}^{n-3}{x}^{n-6}$ ...(i)
Since the fourth term is independent of $x$,

$n-6=0$
$n=6$

Substituting this value in (i), we get

${}^6{C}_3(p{)}^3=\frac{5}{2}$
$\Rightarrow 20p^3=\frac{5}{2}$
$\Rightarrow p^3=\frac{1}{8}$
$\Rightarrow p=\frac{1}{2}$
Therefore $(n,p)=(6,\frac{1}{2})$