To find the terms of a geometric sequence, you can use the formula for the $n$th term of a geometric sequence:

$a_n = a_1 \times r^{(n-1)}$

Given:

• $a_1 = \frac{1}{2}$
• $r = \frac{1}{2}$

Now, let's find the first six terms:

1. $a_1 = \frac{1}{2}$

2. $a_2 = a_1 \times r = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

3. $a_3 = a_1 \times r^2 = \frac{1}{2} \times \left(\frac{1}{2}\right)^2 = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

4. $a_4 = a_1 \times r^3 = \frac{1}{2} \times \left(\frac{1}{2}\right)^3 = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}$

5. $a_5 = a_1 \times r^4 = \frac{1}{2} \times \left(\frac{1}{2}\right)^4 = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32}$

6. $a_6 = a_1 \times r^5 = \frac{1}{2} \times \left(\frac{1}{2}\right)^5 = \frac{1}{2} \times \frac{1}{32} = \frac{1}{64}$

So, the first six terms of the geometric sequence are:

$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}$

Would you like more details or have any questions?

Here are five related questions:

1. What is the 10th term of the sequence?
2. How do you find the sum of the first $n$ terms of a geometric sequence?
3. If the common ratio were negative, how would the sequence change?
4. How does the geometric sequence differ from an arithmetic sequence?
5. What is the sum of an infinite geometric series with $|r| < 1$?

Tip: In a geometric sequence, if the common ratio $r$ is between -1 and 1 (excluding 0), the terms will get closer and closer to 0 as the sequence progresses.