If the coefficient of x in \(\Big(\text{x}^{2}+\frac{\lambda}{\text{x}}\Big)^{5}\) is 270, then \(\lambda=\)

If the coefficient of x in \(\Big(\text{x}^{2}+\frac{\lambda}{\text{x}}\Big)^{5}\) is 270, then \(\lambda=\)

  1. 3

  2. None of these.

  3. 4

  4. 5

Solution

Solution:

The coefficient of x in the given expansion where x occurs at the (r + 1)th term.

We have,

\({^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}\)

\(={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}\)

For it to contain x, we must have

\(10-3\text{r}=1\)

\(\Rightarrow \text{r}=3\)

Coefficient of x in the given expansion,

\(={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}=10\lambda^{3}\)

Now, we have

\(10\lambda^{3}=270\)

\(\Rightarrow \lambda^{3}=27\)

\(\Rightarrow \lambda=3\)

Asked in: RDSHARMA

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