Question

If the coefficient of $x$ in ${(x^2+\frac{A}{x})}^5$ is $270$, then $A=?$

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$3$

Solution:- (A) $3$

As we know that general term in an expnsion ${(a+b)}^n$ is given as-

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Therefore,

In the expansion,

${(x^2+\frac{A}{x})}^5$

$a=x^2$

$b=\frac{A}{x}$

$n=5$

$\therefore T_{r+1}={}^5{C}_r{\left(x^2\right)}^{5-r}{\left(\frac{A}{x}\right)}^r$

$\Rightarrow T_{r+1}={}^5{C}_r{x}^{10-2r}{A}^{-r}$

For coefficient of $x^1$-

$10-3r=1\Rightarrow r=\frac{9}{3}=3$

Therefore,

Coefficient of $x={}^5{C}_3{A}^3=270\left(\text{Given}\right)$

$\therefore {}^5{C}_3{A}^3=270$

$\frac{5\times 4\times 3!}{3!\times 2\times 1}\times A^3=270$

$\Rightarrow A^3=\frac{270}{10}$

$\Rightarrow A=\sqrt[3]{327}=3$

Hence the value of $A$ is $3$.