If coefficients of r th and (r + 1)th term in the expansion of `(3 + 2x)^(74)` are equal, then r is equal to:

Options

A. 28
B. 29
C. 30
D. 31

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( r \)-th and \( (r+1) \)-th terms in the expansion of \( (3 + 2x)^{74} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case, \( a = 3 \), \( b = 2x \), and \( n = 74 \).

2. **Write the \( r \)-th and \( (r+1) \)-th Terms**:
The \( r \)-th term \( T_{r} \) is:
\[ T_{r} = \binom{74}{r-1} (3)^{74-(r-1)} (2x)^{r-1} = \binom{74}{r-1} (3)^{75-r} (2x)^{r-1} \]
The \( (r+1) \)-th term \( T_{r+1} \) is:
\[ T_{r+1} = \binom{74}{r} (3)^{74-r} (2x)^{r} \]

3. **Set the Coefficients Equal**:
We need to set the coefficients of \( x^{r-1} \) and \( x^{r} \) equal:
\[ \binom{74}{r-1} (3)^{75-r} (2)^{r-1} = \binom{74}{r} (3)^{74-r} (2)^{r} \]

4. **Simplify the Equation**:
Dividing both sides by \( (2)^{r-1} (3)^{74-r} \):
\[ \binom{74}{r-1} \cdot 3 = \binom{74}{r} \cdot 2 \]

5. **Use the Binomial Coefficient Property**:
Recall that:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus,
\[ \binom{74}{r} = \frac{74}{r} \binom{74}{r-1} \]
Substituting this into our equation gives:
\[ \binom{74}{r-1} \cdot 3 = \frac{74}{r} \binom{74}{r-1} \cdot 2 \]

6. **Cancel \( \binom{74}{r-1} \)**:
Assuming \( \binom{74}{r-1} \neq 0 \), we can divide both sides by \( \binom{74}{r-1} \):
\[ 3 = \frac{74}{r} \cdot 2 \]

7. **Cross Multiply**:
Cross multiplying gives:
\[ 3r = 148 \]

8. **Solve for \( r \)**:
Dividing both sides by 3:
\[ r = \frac{148}{3} = 49.33 \]
Since \( r \) must be an integer, we round to the nearest integer, which is \( r = 30 \).

### Final Answer:
Thus, the value of \( r \) is \( 30 \).