if the coefficients of `x^(5)" and "x^(15)` in the expansion of `(x^(2)+(a)/(x^(3)))^(10)` are equal then then the positive value of 'a' is:

Options

A. `(1)/(2 sqrt(3))`
B. `(1)/(sqrt(3))`
C. 1
D. `2sqrt(3)`

Answer

To solve the problem, we need to find the positive value of \( a \) such that the coefficients of \( x^5 \) and \( x^{15} \) in the expansion of \( (x^2 + \frac{a}{x^3})^{10} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (x^2 + \frac{a}{x^3})^{10} \) is given by:
\[ T_{r+1} = \binom{10}{r} (x^2)^{10-r} \left(\frac{a}{x^3}\right)^r \]
Simplifying this, we get:
\[ T_{r+1} = \binom{10}{r} a^r x^{20 - 2r - 3r} = \binom{10}{r} a^r x^{20 - 5r} \]

2. **Find the Coefficient of \( x^5 \)**:
For \( x^5 \), we set \( 20 - 5r = 5 \):
\[ 20 - 5r = 5 \implies 5r = 15 \implies r = 3 \]
The coefficient of \( x^5 \) is:
\[ \text{Coefficient of } x^5 = \binom{10}{3} a^3 \]

3. **Find the Coefficient of \( x^{15} \)**:
For \( x^{15} \), we set \( 20 - 5r = 15 \):
\[ 20 - 5r = 15 \implies 5r = 5 \implies r = 1 \]
The coefficient of \( x^{15} \) is:
\[ \text{Coefficient of } x^{15} = \binom{10}{1} a^1 \]

4. **Set the Coefficients Equal**:
Since the coefficients of \( x^5 \) and \( x^{15} \) are equal, we have:
\[ \binom{10}{3} a^3 = \binom{10}{1} a \]

5. **Substitute the Binomial Coefficients**:
We know:
\[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
\[ \binom{10}{1} = 10 \]
Thus, we can rewrite the equation:
\[ 120 a^3 = 10 a \]

6. **Simplify the Equation**:
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 120 a^2 = 10 \implies a^2 = \frac{10}{120} = \frac{1}{12} \]

7. **Solve for \( a \)**:
Taking the square root of both sides, we find:
\[ a = \sqrt{\frac{1}{12}} = \frac{1}{2\sqrt{3}} \]

8. **Final Answer**:
The positive value of \( a \) is:
\[ a = \frac{1}{2\sqrt{3}} \]