The positive value of `'a'` so that the coefficient of `x^5` is equal to that of `x^15` in the expansion `(x^2+a/x^3)^10` is

Options

A. `(1)/(2sqrt(3))`
B. `(1)/(sqrt(3))`
C. `(sqrt(3))/(6)`
D. `(1)/(3)`

Answer

To find the positive value of \( a \) such that the coefficient of \( x^5 \) is equal to the coefficient of \( x^{15} \) in the expansion of \( (x^2 + \frac{a}{x^3})^{10} \), we will follow these steps:

### Step-by-step Solution:

1. **Identify the General Term:**
The general term \( T_r \) in the expansion of \( (x^2 + \frac{a}{x^3})^{10} \) can be expressed as:
\[ T_r = \binom{10}{r} (x^2)^{10-r} \left(\frac{a}{x^3}\right)^r \]
Simplifying this gives:
\[ T_r = \binom{10}{r} a^r x^{20 - 2r - 3r} = \binom{10}{r} a^r x^{20 - 5r} \]

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

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

4. **Set the Coefficients Equal:**
We need to set the coefficients equal to each other:
\[ \binom{10}{3} a^3 = \binom{10}{1} a \]
Substituting the binomial coefficients:
\[ 120 a^3 = 10 a \]

5. **Solve for \( a \):**
Rearranging gives:
\[ 120 a^3 - 10 a = 0 \]
Factoring out \( a \):
\[ a(120 a^2 - 10) = 0 \]
This gives us:
\[ a = 0 \quad \text{or} \quad 120 a^2 = 10 \implies a^2 = \frac{10}{120} = \frac{1}{12} \]
Therefore:
\[ a = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}} \]

### Final Answer:
The positive value of \( a \) is:
\[ \boxed{\frac{1}{2\sqrt{3}}} \]