If the coefficient of x in the expansion of { | vedclass

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Question

(c) ${T_{r + 1}} = {}^5{C_r}{({x^2})^{5 - r}}{\left( {\frac{k}{x}} \right)^r}$

For coefficient of $x$, $10 - 2r - r = 1\,\,\,\, \Rightarrow r = 3$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(c) ${T_{r + 1}} = {}^5{C_r}{({x^2})^{5 - r}}{\left( {\frac{k}{x}} \right)^r}$

For coefficient of $x$, $10 - 2r - r = 1\,\,\,\, \Rightarrow r = 3$

Hence, ${T_{3 + 1}} = {}^5{C_3}{({x^2})^{5 - 3}}{\left( {\frac{k}{x}} \right)^3}$

According to question, $10\,{k^3} = 270\,\, \Rightarrow \,\,k = 3$.

If the coefficient of $x$ in the expansion of ${\left( {{x^2} + \frac{k}{x}} \right)^5}$ is $270$, then $k =$

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