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

Explanation

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 in the expansion of \((x^2 + \frac{a}{x^3})^{10}\) can be expressed using the binomial theorem:
\[ T_k = \binom{10}{k} (x^2)^{10-k} \left(\frac{a}{x^3}\right)^k = \binom{10}{k} a^k x^{20 - 5k} \]
Here, \(T_k\) is the \(k^{th}\) term in the expansion.

2. **Find the Coefficient of \(x^5\)**:
We need \(20 - 5k = 5\):
\[ 20 - 5k = 5 \implies 5k = 15 \implies k = 3 \]
The coefficient of \(x^5\) is:
\[ \text{Coefficient of } x^5 = \binom{10}{3} a^3 \]

3. **Find the Coefficient of \(x^{15}\)**:
We need \(20 - 5k = 15\):
\[ 20 - 5k = 15 \implies 5k = 5 \implies k = 1 \]
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 set them equal to each other:
\[ \binom{10}{3} a^3 = \binom{10}{1} a \]

5. **Calculate the Binomial Coefficients**:
\[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]
\[ \binom{10}{1} = 10 \]
Thus, the equation becomes:
\[ 120 a^3 = 10 a \]

6. **Rearrange the Equation**:
\[ 120 a^3 - 10 a = 0 \]
Factor out \(a\):
\[ a(120 a^2 - 10) = 0 \]

7. **Solve for \(a\)**:
This gives us two cases:
- \(a = 0\) (not a positive value)
- \(120 a^2 - 10 = 0\)
\[ 120 a^2 = 10 \implies a^2 = \frac{10}{120} = \frac{1}{12} \]
\[ a = \sqrt{\frac{1}{12}} = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}} \]

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