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 (a) `23` (b) `24` (c) `25` (d) `26`

Options

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

Explanation

To solve the problem, we need to find the values of \( n \) such that 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 in the expansion**:
The general term \( T_k \) in the expansion of \( \left( a + b \right)^n \) is given by:
\[ T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
For our case, \( a = \frac{3}{2} \) and \( b = \frac{3}{3} = 1 \). The 6th term \( T_6 \) is:
\[ T_6 = \binom{n}{5} \left( \frac{3}{2} \right)^{n-5} \cdot 1^5 = \binom{n}{5} \left( \frac{3}{2} \right)^{n-5} \]

2. **Find the 5th and 4th terms**:
- The 5th term \( T_5 \):
\[ T_5 = \binom{n}{4} \left( \frac{3}{2} \right)^{n-4} \]
- The 4th term \( T_4 \):
\[ T_4 = \binom{n}{3} \left( \frac{3}{2} \right)^{n-3} \]

3. **Set up the conditions for maximum**:
For the 6th term to be numerically greatest, we need:
\[ \frac{T_6}{T_5} > 1 \quad \text{and} \quad \frac{T_6}{T_4} < 1 \]

4. **Calculate \( \frac{T_6}{T_5} \)**:
\[ \frac{T_6}{T_5} = \frac{\binom{n}{5} \left( \frac{3}{2} \right)^{n-5}}{\binom{n}{4} \left( \frac{3}{2} \right)^{n-4}} = \frac{\binom{n}{5}}{\binom{n}{4}} \cdot \frac{1}{\frac{3}{2}} = \frac{n-4}{5} \cdot \frac{2}{3} \]
Setting this greater than 1:
\[ \frac{2(n-4)}{15} > 1 \implies 2(n-4) > 15 \implies n - 4 > 7.5 \implies n > 11.5 \]

5. **Calculate \( \frac{T_6}{T_4} \)**:
\[ \frac{T_6}{T_4} = \frac{\binom{n}{5} \left( \frac{3}{2} \right)^{n-5}}{\binom{n}{3} \left( \frac{3}{2} \right)^{n-3}} = \frac{\binom{n}{5}}{\binom{n}{3}} \cdot \left( \frac{3}{2} \right)^{-2} = \frac{(n-4)(n-5)}{3 \cdot 2} \cdot \frac{4}{9} \]
Setting this less than 1:
\[ \frac{(n-4)(n-5)}{6} < 1 \implies (n-4)(n-5) < 6 \]
This gives us a quadratic inequality:
\[ n^2 - 9n + 20 < 6 \implies n^2 - 9n + 14 < 0 \]
Factoring:
\[ (n-7)(n-2) < 0 \]
The solution to this inequality is:
\[ 2 < n < 7 \]

6. **Combine the inequalities**:
From \( n > 11.5 \) and \( 2 < n < 7 \), we find that \( n \) must be an integer such that:
\[ n = 12, 13 \]

7. **Sum of possible integral values of \( n \)**:
The possible integral values of \( n \) are 12 and 13. Therefore, the sum is:
\[ 12 + 13 = 25 \]

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