If the coefficient of x in

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

$10 \lambda^{3}=270$

$\Rightarrow \lambda^{3}=27$

$\Rightarrow \lambda=3$

 

 

 

 

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