If `a_(n)` represents the nth term of the sequence `45, 18, 6, …, and a_(1)` represents the first term, then `a_(n)=`

Options

A. `6((1)/(3))^(n)`
B. `6((1)/(3))^(n-1)`
C. `54((1)/(3))^(n)`
D. `54((1)/(3))^(n-1)`

Answer

To find the nth term of the sequence \( a_n \) represented by the terms \( 45, 18, 6, \ldots \), we will analyze the relationship between the terms step by step.

### Step 1: Identify the first few terms of the sequence
The first three terms of the sequence are:
- \( a_1 = 45 \)
- \( a_2 = 18 \)
- \( a_3 = 6 \)

### Step 2: Find the relationship between the terms
Let's look at how each term relates to the previous term:
- To get from \( a_1 \) to \( a_2 \):
\[ a_2 = \frac{a_1}{2.5} = \frac{45}{2.5} = 18 \]
- To get from \( a_2 \) to \( a_3 \):
\[ a_3 = \frac{a_2}{3} = \frac{18}{3} = 6 \]

### Step 3: Identify a pattern
From the calculations, we can see that:
- \( a_2 = \frac{a_1}{2.5} \)
- \( a_3 = \frac{a_2}{3} \)

Now, let's express \( a_n \) in terms of \( a_1 \):
- We can express \( a_n \) as:
\[ a_n = a_1 \cdot \left(\frac{1}{k}\right)^{n-1} \]
Where \( k \) is the factor by which each term is divided to get the next term.

### Step 4: Find the general formula
From the sequence, we can observe that:
- \( a_2 = a_1 \cdot \frac{2}{5} \)
- \( a_3 = a_2 \cdot \frac{2}{3} = a_1 \cdot \frac{2}{5} \cdot \frac{1}{3} \)

Thus, we can generalize this to:
\[ a_n = 45 \cdot \left(\frac{2}{5}\right)^{n-1} \cdot \left(\frac{1}{3}\right)^{n-2} \]

### Step 5: Simplify the formula
To simplify, we can express it as:
\[ a_n = 45 \cdot \frac{2^{n-1}}{5^{n-1} \cdot 3^{n-2}} \]
This can be further simplified to:
\[ a_n = 45 \cdot \frac{2^{n-1}}{5^{n-1} \cdot 3^{n-2}} = 45 \cdot \frac{2^{n-1}}{5^{n-1} \cdot 3^{n-2}} = 45 \cdot \frac{2^{n-1}}{15^{n-1}} \]

### Final formula
Thus, the nth term of the sequence can be expressed as:
\[ a_n = 45 \cdot \left(\frac{2}{15}\right)^{n-1} \]