The problem asks to find the 9th term of the sequence, where the first three terms are given as:

$32, 16, 8, \dots$

We can observe that this is a geometric sequence because each term is half of the previous one.

Step-by-Step Solution:

1. Identify the common ratio:
   To find the common ratio $r$, divide the second term by the first term: $r = \frac{16}{32} = \frac{1}{2}$

2. Formula for the $n$-th term of a geometric sequence:
   The general formula for the $n$-th term of a geometric sequence is: $a_n = a_1 \times r^{n-1}$ Where:

   • $a_1$ is the first term,
   • $r$ is the common ratio,
   • $n$ is the term number.

3. Substitute the known values:
   In this case:

   • $a_1 = 32$,
   • $r = \frac{1}{2}$,
   • $n = 9$.

   Plug these values into the formula: $a_9 = 32 \times \left(\frac{1}{2}\right)^{9-1} = 32 \times \left(\frac{1}{2}\right)^8$

4. Simplify: $\left(\frac{1}{2}\right)^8 = \frac{1}{256}$ So: $a_9 = 32 \times \frac{1}{256} = \frac{32}{256} = \frac{1}{8}$

Therefore, the 9th term is $\frac{1}{8}$ or 0.125.

Would you like further details on this?
Here are some related questions for practice:

1. What is the 10th term of this sequence?
2. How would you find the sum of the first 9 terms of this geometric sequence?
3. What happens to the sequence as $n$ approaches infinity?
4. Can you write the recursive formula for this geometric sequence?
5. How would the sequence change if the common ratio was negative?

Tip: Geometric sequences decrease or grow exponentially depending on the common ratio. Always check the ratio carefully!