Which term of the progression `18 ,-12 ,8, i s(512)/(729)?`

Options

A. `7^(th)`
B. `9^(th)`
C. `11^(th)`
D. `13^(th)`

Answer

To find which term of the progression \( 18, -12, 8, \ldots \) is \( \frac{512}{729} \), we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \( A \) of the progression is:
\[ A = 18 \]
The second term is \( -12 \), and the third term is \( 8 \). To find the common ratio \( R \), we can use the formula:
\[ R = \frac{\text{second term}}{\text{first term}} = \frac{-12}{18} = -\frac{2}{3} \]

### Step 2: Write the formula for the \( n \)-th term
The formula for the \( n \)-th term of a geometric progression is given by:
\[ a_n = A \cdot R^{n-1} \]
Substituting the values of \( A \) and \( R \):
\[ a_n = 18 \cdot \left(-\frac{2}{3}\right)^{n-1} \]

### Step 3: Set the \( n \)-th term equal to \( \frac{512}{729} \)
We need to find \( n \) such that:
\[ 18 \cdot \left(-\frac{2}{3}\right)^{n-1} = \frac{512}{729} \]

### Step 4: Isolate the common ratio term
Dividing both sides by 18:
\[ \left(-\frac{2}{3}\right)^{n-1} = \frac{512}{729 \cdot 18} \]
Calculating the right-hand side:
\[ 729 \cdot 18 = 13122 \quad \text{(since } 729 = 3^6 \text{ and } 18 = 2 \cdot 3^2\text{)} \]
Thus,
\[ \frac{512}{13122} \]

### Step 5: Simplify \( \frac{512}{13122} \)
Now, we can simplify \( \frac{512}{13122} \):
\[ 512 = 2^9 \quad \text{and} \quad 13122 = 2 \cdot 3^{6} \cdot 2 = 2^1 \cdot 3^6 \]
So,
\[ \frac{512}{13122} = \frac{2^9}{2^1 \cdot 3^6} = \frac{2^{9-1}}{3^6} = \frac{2^8}{3^6} = \frac{256}{729} \]

### Step 6: Set the equation for the powers
Now we have:
\[ \left(-\frac{2}{3}\right)^{n-1} = \frac{256}{729} \]
This can be rewritten as:
\[ \left(-\frac{2}{3}\right)^{n-1} = \left(-\frac{2}{3}\right)^{8} \]
Thus, we can equate the exponents:
\[ n - 1 = 8 \]
So,
\[ n = 9 \]

### Final Answer
The term \( \frac{512}{729} \) is the 9th term of the progression.

---