Question:
If the coefficient of $x$ in $\left(x^{2}+\frac{\lambda}{x}\right)^{5}$ is 270, then $\lambda=$
(a) 3
(b) 4
(c) 5
(d) none of these
Solution:
(a) 3
The coefficient of $x$ in the given expansion where $x$ occurs at the $(r+1)$ th term.
We have:
${ }^{5} C_{r}\left(x^{2}\right)^{5-r}\left(\frac{\lambda}{x}\right)^{r}$
$={ }^{5} C_{r} \lambda^{r} x^{10-2 r-r}$
For it to contain $x$, we must have :
$10-3 r=1$
$\Rightarrow r=3$
$\therefore$ Coefficient of $x$ in the given expansion:
${ }^{5} C_{3} \lambda^{3}=10 \lambda^{3}$
Now, we have
$10 \lambda^{3}=270$
$\Rightarrow \lambda^{3}=27$
$\Rightarrow \lambda=3$