If `6^(th)` term in the expansion of `((3)/(2)+(x)/(3))^(n)` is numerically greatest, when `x=3`, then the sum of possible integral values of `'n'` is

Options

A. `23`
B. `24`
C. `25`
D. `26`

Explanation

To solve the problem, we need to find the integral values of \( n \) for which the 6th term in the expansion of \( \left(\frac{3}{2} + \frac{x}{3}\right)^n \) is numerically greatest when \( x = 3 \).

### Step-by-Step Solution:

1. **Identify the 6th Term**:
The \( k \)-th term in the binomial expansion is given by:
\[ T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
Here, \( a = \frac{3}{2} \) and \( b = \frac{x}{3} \). For the 6th term, \( k = 6 \):
\[ T_6 = \binom{n}{5} \left(\frac{3}{2}\right)^{n-5} \left(\frac{x}{3}\right)^5 \]

2. **Substituting \( x = 3 \)**:
\[ T_6 = \binom{n}{5} \left(\frac{3}{2}\right)^{n-5} \left(1\right)^5 = \binom{n}{5} \left(\frac{3}{2}\right)^{n-5} \]

3. **Condition for Greatest Term**:
For the 6th term to be the greatest, we need:
\[ \frac{T_5}{T_4} > 1 \quad \text{and} \quad \frac{T_6}{T_5} < 1 \]

4. **Calculate \( T_5 \) and \( T_4 \)**:
- **5th Term**:
\[ T_5 = \binom{n}{4} \left(\frac{3}{2}\right)^{n-4} \]
- **4th Term**:
\[ T_4 = \binom{n}{3} \left(\frac{3}{2}\right)^{n-3} \]

5. **Set Up the Inequality**:
\[ \frac{T_5}{T_4} = \frac{\binom{n}{4} \left(\frac{3}{2}\right)^{n-4}}{\binom{n}{3} \left(\frac{3}{2}\right)^{n-3}} > 1 \]
Simplifying gives:
\[ \frac{\binom{n}{4}}{\binom{n}{3}} \cdot \frac{1}{\frac{3}{2}} > 1 \]
\[ \frac{n-4}{5} \cdot \frac{2}{3} > 1 \]
\[ n - 4 > \frac{15}{2} \quad \Rightarrow \quad n > \frac{23}{2} \quad \Rightarrow \quad n > 11.5 \]

6. **Second Condition**:
Now for \( \frac{T_6}{T_5} < 1 \):
\[ \frac{T_6}{T_5} = \frac{\binom{n}{5} \left(\frac{3}{2}\right)^{n-5}}{\binom{n}{4} \left(\frac{3}{2}\right)^{n-4}} < 1 \]
Simplifying gives:
\[ \frac{\binom{n}{5}}{\binom{n}{4}} \cdot \frac{1}{\frac{3}{2}} < 1 \]
\[ \frac{n-5}{6} \cdot \frac{2}{3} < 1 \]
\[ n - 5 < 9 \quad \Rightarrow \quad n < 14 \]

7. **Combine Results**:
From the inequalities, we have:
\[ 11.5 < n < 14 \]
The possible integral values for \( n \) are \( 12 \) and \( 13 \).

8. **Sum of Possible Values**:
\[ 12 + 13 = 25 \]

### Final Answer:
The sum of possible integral values of \( n \) is \( \boxed{25} \).