If the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of `(sqrt2+(1)/(sqrt3))^x` is `1/6`, then x is

Options

A. 3
B. 6
C. 10
D. 9

Answer

To solve the problem, we need to find the value of \( x \) such that the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of \( \left( \sqrt{2} + \frac{1}{\sqrt{3}} \right)^x \) is \( \frac{1}{6} \).

### Step-by-Step Solution:

1. **Identify the Terms**:
- The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
- Here, \( a = \sqrt{2} \), \( b = \frac{1}{\sqrt{3}} \), and \( n = x \).

2. **Find the 7th Term from the Beginning**:
- The 7th term from the beginning corresponds to \( r = 7 \) (i.e., \( r-1 = 6 \)):
\[ T_7 = \binom{x}{6} (\sqrt{2})^{x-6} \left(\frac{1}{\sqrt{3}}\right)^{6} \]
- Simplifying this:
\[ T_7 = \binom{x}{6} (\sqrt{2})^{x-6} \cdot \frac{1}{3^{3}} = \binom{x}{6} (\sqrt{2})^{x-6} \cdot \frac{1}{27} \]

3. **Find the 7th Term from the End**:
- The 7th term from the end corresponds to \( r = x - 6 \) (i.e., \( n - r + 1 = 7 \)):
\[ T_{x-6} = \binom{x}{x-6} (\sqrt{2})^{6} \left(\frac{1}{\sqrt{3}}\right)^{x-6} \]
- Simplifying this:
\[ T_{x-6} = \binom{x}{6} (\sqrt{2})^{6} \cdot \left(\frac{1}{\sqrt{3}}\right)^{x-6} = \binom{x}{6} \cdot 2^{3} \cdot \frac{1}{3^{(x-6)/2}} \]

4. **Set Up the Ratio**:
- We know the ratio of the 7th term from the beginning to the 7th term from the end is given as:
\[ \frac{T_7}{T_{x-6}} = \frac{1}{6} \]
- Substituting the expressions we found:
\[ \frac{\binom{x}{6} (\sqrt{2})^{x-6} \cdot \frac{1}{27}}{\binom{x}{6} \cdot 2^{3} \cdot \frac{1}{3^{(x-6)/2}}} = \frac{1}{6} \]
- The \( \binom{x}{6} \) cancels out:
\[ \frac{(\sqrt{2})^{x-6}}{27 \cdot 8 \cdot 3^{(6-x)/2}} = \frac{1}{6} \]

5. **Simplify the Equation**:
- Rearranging gives:
\[ \frac{(\sqrt{2})^{x-6}}{216 \cdot 3^{(6-x)/2}} = \frac{1}{6} \]
- Cross-multiplying:
\[ (\sqrt{2})^{x-6} = 36 \cdot 3^{(6-x)/2} \]

6. **Equate the Powers**:
- Taking logarithms or equating powers gives:
\[ 2^{(x-6)/2} = 6^{2} \cdot 3^{(6-x)/2} \]
- This leads to:
\[ 2^{(x-6)/2} = 2^{4} \cdot 3^{(6-x)/2} \]

7. **Solve for \( x \)**:
- Equating the exponents:
\[ \frac{x-6}{2} = 4 + \frac{6-x}{2} \]
- Simplifying gives:
\[ x - 6 = 8 + 6 - x \]
\[ 2x = 20 \implies x = 10 \]

### Final Answer:
The value of \( x \) is \( 10 \).