Question

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

Answer 1

${(ax+\frac{1}{x})}^n$

${=}^n{C}_0{\left(ax\right)}^n{+}^n{C}_1{\left(ax\right)}^{n-1}\left(\frac{1}{x}\right){+}^n{C}_2{\left(ax\right)}^{n-2}{\left(\frac{1}{x}\right)}^2{+}^n{C}_3{\left(ax\right)}^{n-3}{\left(\frac{1}{x}\right)}^3+...$

Fourth term$=t_4{=}^n{C}_3{\left(ax\right)}^{n-3}{\left(\frac{1}{x}\right)}^3$

$\Rightarrow t_4{=}^n{C}_3{\left(ax\right)}^{n-3}\left(\frac{1}{x^3}\right)$

${\Rightarrow }^n{C}_3{a}^{n-3}{x}^{n-6}=\frac{5}{2}{x}^0$

Comparing the coefficeint of $x$

$\Rightarrow n-6=0$

$\Rightarrow n=6$

So,${}_{C_3}^n{a}^{n-3}=\frac{5}{2}$

${\Rightarrow }_{C_3}^6{a}^{6-3}=\frac{5}{2}$

${\Rightarrow }_{C_3}^6{a}^3=\frac{5}{2}$

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

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

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

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

$\therefore n=6$ and $a=\frac{1}{2}$