If the coefficients of `x^2a n d\ x^3` in the expansion o `(3+a x)^9` are the same, then the value of `a` is `-7/9` b. `-9/7` c. `7/9` d. `9/7`

Explanation

To solve the problem, we need to find the value of \( a \) such that the coefficients of \( x^2 \) and \( x^3 \) in the expansion of \( (3 + ax)^9 \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the expansion of \( (p + q)^n \) is given by:
\[ T_r = \binom{n}{r} p^{n-r} q^r \]
For our case, \( p = 3 \), \( q = ax \), and \( n = 9 \). Thus, the general term becomes:
\[ T_r = \binom{9}{r} (3)^{9-r} (ax)^r = \binom{9}{r} 3^{9-r} a^r x^r \]

2. **Find the Coefficient of \( x^2 \)**:
To find the coefficient of \( x^2 \), we set \( r = 2 \):
\[ \text{Coefficient of } x^2 = \binom{9}{2} 3^{9-2} a^2 = \binom{9}{2} 3^7 a^2 \]
Calculate \( \binom{9}{2} \):
\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \]
Therefore, the coefficient of \( x^2 \) is:
\[ 36 \cdot 3^7 \cdot a^2 \]

3. **Find the Coefficient of \( x^3 \)**:
To find the coefficient of \( x^3 \), we set \( r = 3 \):
\[ \text{Coefficient of } x^3 = \binom{9}{3} 3^{9-3} a^3 = \binom{9}{3} 3^6 a^3 \]
Calculate \( \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
Therefore, the coefficient of \( x^3 \) is:
\[ 84 \cdot 3^6 \cdot a^3 \]

4. **Set the Coefficients Equal**:
Since the coefficients of \( x^2 \) and \( x^3 \) are equal:
\[ 36 \cdot 3^7 \cdot a^2 = 84 \cdot 3^6 \cdot a^3 \]

5. **Simplify the Equation**:
Divide both sides by \( 3^6 \):
\[ 36 \cdot 3 \cdot a^2 = 84 \cdot a^3 \]
Simplifying gives:
\[ 108 a^2 = 84 a^3 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 84 a^3 - 108 a^2 = 0 \]
Factor out \( a^2 \):
\[ a^2 (84 a - 108) = 0 \]

7. **Solve for \( a \)**:
This gives us two solutions:
\[ a^2 = 0 \quad \text{or} \quad 84 a - 108 = 0 \]
The first solution \( a = 0 \) is trivial. For the second:
\[ 84 a = 108 \implies a = \frac{108}{84} = \frac{9}{7} \]

### Conclusion:
The value of \( a \) is \( \frac{9}{7} \).