The coefficient of x in `(x^2 + a/x)^5` is 270 then a =

Options

A. 3
B. 4
C. 5
D. 2

Answer

To solve the problem of finding the value of \( a \) in the expression \( (x^2 + \frac{a}{x})^5 \) such that the coefficient of \( x \) is 270, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (x^2 + \frac{a}{x})^5 \) can be expressed as:
\[ T_{r+1} = \binom{5}{r} (x^2)^{5-r} \left(\frac{a}{x}\right)^r \]

2. **Simplify the General Term**:
Simplifying the general term:
\[ T_{r+1} = \binom{5}{r} (x^{10 - 2r}) \left(\frac{a^r}{x^r}\right) = \binom{5}{r} a^r x^{10 - 3r} \]

3. **Determine the Condition for Coefficient of \( x \)**:
We need the power of \( x \) to be 1:
\[ 10 - 3r = 1 \]
Solving for \( r \):
\[ 10 - 1 = 3r \implies 3r = 9 \implies r = 3 \]

4. **Substitute \( r \) Back into the General Term**:
Now substitute \( r = 3 \) into the general term:
\[ T_{4} = \binom{5}{3} a^3 x^{10 - 9} = \binom{5}{3} a^3 x^1 \]

5. **Calculate the Coefficient**:
The coefficient of \( x \) is:
\[ \binom{5}{3} a^3 \]
We know that this coefficient equals 270:
\[ \binom{5}{3} a^3 = 270 \]

6. **Calculate \( \binom{5}{3} \)**:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Therefore, we have:
\[ 10 a^3 = 270 \]

7. **Solve for \( a^3 \)**:
Dividing both sides by 10:
\[ a^3 = \frac{270}{10} = 27 \]

8. **Find \( a \)**:
Taking the cube root:
\[ a = \sqrt[3]{27} = 3 \]

### Final Answer:
Thus, the value of \( a \) is:
\[ \boxed{3} \]