The first three terms of a sequence are [tex]\frac{3}{2}, 3, \text{ and } 6[/tex]. What is the [tex]n^{\text{th}}[/tex] term?

1. [tex]3 \cdot 2^n[/tex]
2. [tex]2^{n-3}[/tex]
3. [tex]3 \cdot 2^{n-2}[/tex]
4. [tex]6 \cdot 2^{1-n}[/tex]

< previous (2) (3) confirm (4) next >
Terminate the exam.



Answer :

To determine the [tex]\( n \)[/tex]-th term of the given sequence [tex]\( \frac{3}{2}, 3, 6 \)[/tex], we need to discern the pattern within the sequence.

### Step-by-Step Analysis

1. Identify the Pattern:
- Start by checking if the sequence follows a specific pattern, such as arithmetic or geometric. An arithmetic sequence has a constant difference between terms, while a geometric sequence has a constant ratio.

2. Compute Ratios:
- Calculate the ratio between consecutive terms:
- From [tex]\( \frac{3}{2} \)[/tex] to [tex]\( 3 \)[/tex]:
[tex]\[ \text{Ratio}_1 = \frac{3}{\frac{3}{2}} = 3 \times \frac{2}{3} = 2 \][/tex]
- From [tex]\( 3 \)[/tex] to [tex]\( 6 \)[/tex]:
[tex]\[ \text{Ratio}_2 = \frac{6}{3} = 2 \][/tex]
- Since both ratios are equal ([tex]\( \text{Ratio}_1 = \text{Ratio}_2 = 2 \)[/tex]), the sequence is a geometric sequence with a common ratio ([tex]\( r \)[/tex]) of 2.

3. General Formula for the nth Term:
- The [tex]\( n \)[/tex]-th term of a geometric sequence is given by:
[tex]\[ a \cdot r^{n-1} \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio.

4. Substitute Values:
- Here, [tex]\( a = \frac{3}{2} \)[/tex] and [tex]\( r = 2 \)[/tex].
[tex]\[ \text{nth term} = \frac{3}{2} \cdot 2^{n-1} \][/tex]

### Match with Given Options:

1. Option 1: [tex]\( 3 \cdot 2^n \)[/tex]
- [tex]\( \frac{3}{2} \cdot 2^{n-1} \neq 3 \cdot 2^n \)[/tex]

2. Option 2: [tex]\( 2^{n-3} \)[/tex]
- [tex]\( \frac{3}{2} \cdot 2^{n-1} \neq 2^{n-3} \)[/tex]

3. Option 3: [tex]\( 3 \cdot 2^{n-2} \)[/tex]
- Simplify to see if it matches:
[tex]\[ \frac{3}{2} \cdot 2^{n-1} = 3 \cdot 2^{n-2} \][/tex]
This is correct because:
[tex]\[ \frac{3}{2} \cdot 2^{n-1} = 3 \cdot 2^{n-2} \quad \text{(since } 2^{n-1-1} = 2^{n-2} \text{)} \][/tex]

4. Option 4: [tex]\( 6 \cdot 2^{1-n} \)[/tex]
- [tex]\( \frac{3}{2} \cdot 2^{n-1} \neq 6 \cdot 2^{1-n} \)[/tex]

Therefore, the correct expression for the [tex]\( n \)[/tex]-th term of the sequence is:

Option 3: [tex]\( 3 \cdot 2^{n-2} \)[/tex]