If the coefficient of `x^(2)` and `x^(3)` in the expansion of `(3+ax)^(11)` are equal then a=_______

Answer

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 \( (3 + ax)^{11} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (3 + ax)^{11} \) is given by:
\[ T_{r+1} = \binom{11}{r} \cdot (3)^{11-r} \cdot (ax)^r = \binom{11}{r} \cdot 3^{11-r} \cdot a^r \cdot x^r \]

2. **Find the Coefficient of \( x^2 \)**:
To find the coefficient of \( x^2 \), set \( r = 2 \):
\[ \text{Coefficient of } x^2 = \binom{11}{2} \cdot 3^{11-2} \cdot a^2 = \binom{11}{2} \cdot 3^9 \cdot a^2 \]

3. **Find the Coefficient of \( x^3 \)**:
To find the coefficient of \( x^3 \), set \( r = 3 \):
\[ \text{Coefficient of } x^3 = \binom{11}{3} \cdot 3^{11-3} \cdot a^3 = \binom{11}{3} \cdot 3^8 \cdot a^3 \]

4. **Set the Coefficients Equal**:
Since the coefficients of \( x^2 \) and \( x^3 \) are equal, we can set up the equation:
\[ \binom{11}{2} \cdot 3^9 \cdot a^2 = \binom{11}{3} \cdot 3^8 \cdot a^3 \]

5. **Simplify the Equation**:
Dividing both sides by \( 3^8 \) gives:
\[ \binom{11}{2} \cdot 3 \cdot a^2 = \binom{11}{3} \cdot a^3 \]
Rearranging gives:
\[ \binom{11}{2} \cdot 3 = \binom{11}{3} \cdot a \]

6. **Calculate the Binomial Coefficients**:
- \( \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55 \)
- \( \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \)

Substituting these values into the equation gives:
\[ 55 \cdot 3 = 165 \cdot a \]

7. **Solve for \( a \)**:
\[ 165a = 165 \implies a = 1 \]

Thus, the value of \( a \) is:

\[ \boxed{1} \]