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If the coefficient of $x$ is in expansion of $\left(x^2+\frac{k}{x}\right)^5$ is 270 , then $k$ is equal to
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Verified Answer
The correct answer is:
$3$
General term in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is
$$
\begin{aligned}
T_{r+1} & ={ }^5 C_r\left(x^2\right)^{5-r}\left(\frac{k}{x}\right)^r \\
& ={ }^5 C_r x^{10-3 r} \cdot k^r
\end{aligned}
$$
Let this term contains $x$ then, $10-3 r=1 \Rightarrow 3 r=9 \Rightarrow r=3$, then coefficient of $x={ }^5 C_3 k^3=10 k^3$ Given that $10 k^3=270$
$$
\therefore \quad k^3=27 \Rightarrow k=3
$$
$$
\begin{aligned}
T_{r+1} & ={ }^5 C_r\left(x^2\right)^{5-r}\left(\frac{k}{x}\right)^r \\
& ={ }^5 C_r x^{10-3 r} \cdot k^r
\end{aligned}
$$
Let this term contains $x$ then, $10-3 r=1 \Rightarrow 3 r=9 \Rightarrow r=3$, then coefficient of $x={ }^5 C_3 k^3=10 k^3$ Given that $10 k^3=270$
$$
\therefore \quad k^3=27 \Rightarrow k=3
$$
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