A. Multiple Choice. Read and analyze the following items and determine the letter 
of the correct answer from the given choices. Write your answer on a separate 
sheet of paper.
1. What expression is same as (𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎)(𝑏 ∙ 𝑏)?
A. (4𝑎)(2𝑏) C. a4 b2
B. a2 b2 D. 8ab
2. What is the common ratio of the sequence 36, 12, 4, …?
A. 3 C. 1
3
B. – 3 D. –
1
3
3. The first term of a geometric sequence is 5, and the second term is 10, what is 
the sixth term?
A. 160 C. 320
B. 80 D. 640
4. The common ratio of a geometric sequence is 1
2
, and the fourth term is 1
4
. What 
is the third term of the sequence?
A. 1 C. 1
4
B. 1
2
D. 1
8
5. The common ratio of a geometric sequence is 3
2
. Find the first three 
 terms if the fifth term is 1. 
A. 16
81
, 
8
27
, 
4
9
C. 32
243
, 
16
81
, 
8
27
B. 8
27
, 
4
9
, 
2
3
D. 2, 1, 1
2

Let's go through each question step-by-step to find the correct answers:

### 1. What expression is the same as \((a \cdot a \cdot a \cdot a)(b \cdot b)\)?

Given: \((a \cdot a \cdot a \cdot a)(b \cdot b)\)

Simplifying:
\[
(a \cdot a \cdot a \cdot a) = a^4, \quad (b \cdot b) = b^2
\]
Therefore, the expression is:
\[
a^4 b^2
\]

**Answer: C. \(a^4 b^2\)**

### 2. What is the common ratio of the sequence \(36, 12, 4, \ldots\)?

To find the common ratio \(r\) of a geometric sequence, divide the second term by the first term:
\[
r = \frac{12}{36} = \frac{1}{3}
\]

**Answer: C. \(\frac{1}{3}\)**

### 3. The first term of a geometric sequence is 5, and the second term is 10. What is the sixth term?

Given a geometric sequence where the first term \(a_1 = 5\) and the second term \(a_2 = 10\), we can find the common ratio \(r\):
\[
r = \frac{a_2}{a_1} = \frac{10}{5} = 2
\]

The \(n\)-th term of a geometric sequence is given by:
\[
a_n = a_1 \cdot r^{n-1}
\]

For the sixth term (\(n = 6\)):
\[
a_6 = 5 \cdot 2^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160
\]

**Answer: A. 160**

### 4. The common ratio of a geometric sequence is \(\frac{1}{2}\), and the fourth term is \(\frac{1}{4}\). What is the third term of the sequence?

Let the third term be \(a_3\). Since the common ratio \(r = \frac{1}{2}\), the fourth term can be expressed as:
\[
a_4 = a_3 \cdot \frac{1}{2} = \frac{1}{4}
\]

Solving for \(a_3\):
\[
a_3 = \frac{1}{4} \cdot 2 = \frac{1}{2}
\]

**Answer: B. \(\frac{1}{2}\)**

### 5. The common ratio of a geometric sequence is \(\frac{3}{2}\). Find the first three terms if the fifth term is 1.

Let the first term be \(a_1 = a\). The \(n\)-th term of a geometric sequence is given by:
\[
a_n = a_1 \cdot r^{n-1}
\]

For the fifth term:
\[
a_5 = a \cdot \left(\frac{3}{2}\right)^4 = 1
\]

Solving for \(a\):
\[
a \cdot \frac{81}{16} = 1 \implies a = \frac{16}{81}
\]

Now, the first three terms are:
\[
a_1 = \frac{16}{81}, \quad a_2 = \frac{16}{81} \cdot \frac{3}{2} = \frac{8}{27}, \quad a_3 = \frac{8}{27} \cdot \frac{3}{2} = \frac{4}{9}
\]

**Answer: A. \(\frac{16}{81}, \frac{8}{27}, \frac{4}{9}\)**

### Final Answers:
1. **C. \(a^4 b^2\)**
2. **C. \(\frac{1}{3}\)**
3. **A. 160**
4. **B. \(\frac{1}{2}\)**
5. **A. \(\frac{16}{81}, \frac{8}{27}, \frac{4}{9}\)**

Would you like more details on any of the solutions or have any questions?

### Related Questions:
1. How do you derive the general formula for the \(n\)-th term of a geometric sequence?
2. What are the properties of a geometric sequence with a negative common ratio?
3. How do you find the sum of a finite geometric series?
4. How do you convert a repeating decimal into a fraction using geometric series?
5. What are the conditions for convergence of an infinite geometric series?

**Tip:** For geometric sequences, always check the sign and magnitude of the common ratio to determine the nature of the sequence (increasing, decreasing, or alternating).

Explore solutions to geometric sequence problems including finding common ratios and calculating specific terms. Includes step-by-step solutions for high school students.

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