Question

If the $4^{th}$ term in the expansion of ${\left[ax+\left(\frac{1}{x}\right)\right]}^n$ is $\frac{5}{2}$, for all $x$ belongs to$R$, then the values of $a$ and $n$ are

• A. $\frac{1}{2},6$
• B. $6,\frac{1}{2}$
• C. $2,6$
• D. None of these

Answer 1

The correct option is A

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

Explanation for Correct Option:

Step 1: Apply the general term ${\mathit{T}}_{\mathbf{r}\mathbf{+}\mathbf{1}}{\mathbf{=}}^{\mathbf{n}}{\mathit{C}}_{\mathbf{r}}\mathbf{\times }{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{r}}\mathbf{\times }{\mathit{y}}^{\mathbf{r}}$

The general term in the expansion of ${(x+y)}^n$is the ${(r+1)}^{th}$ term that can be represented as $T_{r+1}$

Given: ${\left[ax+\left(\frac{1}{x}\right)\right]}^n$

$$\begin{gathered} \therefore {\mathrm{T}}_4=\frac{5}{2} \\ \Rightarrow {}^{\mathrm{n}}\mathrm{C}_3{\left(\mathrm{ax}\right)}^{\mathrm{n}-3}{\left(\frac{1}{\mathrm{x}}\right)}^3=\frac{5}{2} \\ \Rightarrow {}^{\mathrm{n}}\mathrm{C}_3{\left(\mathrm{a}\right)}^{\mathrm{n}-3}{\left(\mathrm{x}\right)}^{\mathrm{n}-3-3}=\frac{5}{2} \\ \Rightarrow {}^{\mathrm{n}}\mathrm{C}_3{\left(\mathrm{a}\right)}^{\mathrm{n}-3}{\left(\mathrm{x}\right)}^{\mathrm{n}-6}=\frac{5}{2} \end{gathered}$$

Step 2: Finding the value of $\mathit{a}$

Since ${\mathrm{T}}_4$ is independent of x

$$\begin{gathered} {\Rightarrow }^6{\mathrm{C}}_3{\mathrm{a}}^3=\frac{5}{2}\left[\because \mathrm{n}-6=0 \\ \Rightarrow \mathrm{n}=6\right] \\ \Rightarrow {\mathrm{a}}^3=\frac{\frac{5}{2}}{{}^6{\mathrm{C}}_3} \\ \Rightarrow {\mathrm{a}}^3=\frac{5}{2}\times \frac{1.3.2.1}{6.5.4} \\ \Rightarrow {\mathrm{a}}^3=\frac{5}{2}\times \frac{6}{6.5.4} \\ \Rightarrow {\mathrm{a}}^3=\frac{1}{8} \\ \Rightarrow \mathrm{a}=\frac{1}{2} \end{gathered}$$

Value of $n=6\mathrm{and}a=\frac{1}{2}$

Hence Option (1) is Correct