QUESTION 3 CW 3.1 If $$ r=\frac{1}{2} $$ and $$ a=3 $$. Which term of the sequence will have a value of $$ \frac{3}{128} $$ ? 3.2 The first three terms of a geometric sequence are $$ x ; y ;-2 y-x $$; Determine the numerical value of $$ \frac{x}{y} $$ 3.3 The information below is that of the convergent geometric series: Determine the value of: 3.3.1 The first term of the convergent geometric series. 3.3.2 The common ratio, $$ r $$, of the convergent geometric series.

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Sure, let's address each part of your question step by step.

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### **Question 3.1**

**Given:**
- Common ratio, $$ r = \frac{1}{2} $$
- First term, $$ a = 3 $$
- We need to find the term number $$ n $$ where the term's value is $$ \frac{3}{128} $$.

**Solution:**

The $$ n $$-th term of a geometric sequence is given by:
$$
a_n = a \cdot r^{n-1}
$$

Set $$ a_n = \frac{3}{128} $$ and solve for $$ n $$:
$$
3 \cdot \left(\frac{1}{2}\right)^{n-1} = \frac{3}{128}
$$

Divide both sides by 3:
$$
\left(\frac{1}{2}\right)^{n-1} = \frac{1}{128}
$$

Recognize that $$ 128 = 2^7 $$:
$$
\left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^7
$$

Since the bases are equal:
$$
n - 1 = 7 \implies n = 8
$$

**Answer:**  
The 8th term of the sequence is $$ \frac{3}{128} $$.

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### **Question 3.2**

**Given:**
- First three terms of a geometric sequence: $$ x $$, $$ y $$, and $$ -2y - x $$
- We need to find the numerical value of $$ \frac{x}{y} $$.

**Solution:**

In a geometric sequence, the ratio between consecutive terms is constant. Let’s denote the common ratio as $$ r $$.

So:
$$
r = \frac{y}{x} = \frac{-2y - x}{y}
$$

Set up the equation:
$$
\frac{y}{x} = \frac{-2y - x}{y}
$$

Multiply both sides by $$ y $$ to eliminate the denominator:
$$
y^2 = (-2y - x) \cdot x
$$

But since $$ \frac{y}{x} = r $$, let’s express $$ y = r x $$.

Substitute $$ y = r x $$ into the equation:
$$
(r x)^2 = (-2 r x - x) \cdot x
$$
$$
r^2 x^2 = (-2r - 1) x^2
$$

Divide both sides by $$ x^2 $$ (assuming $$ x \neq 0 $$):
$$
r^2 = -2r - 1
$$

Rearrange the equation:
$$
r^2 + 2r + 1 = 0
$$

Factor the quadratic:
$$
(r + 1)^2 = 0 \implies r = -1
$$

Since $$ r = \frac{y}{x} = -1 $$, it follows that:
$$
\frac{x}{y} = -1
$$

**Answer:**  
The numerical value of $$ \frac{x}{y} $$ is **-1**.

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### **Question 3.3**

**Given:**
- Information about a convergent geometric series.

**Problem:**
- **3.3.1** Determine the first term of the convergent geometric series.
- **3.3.2** Determine the common ratio, $$ r $$, of the convergent geometric series.

**Solution:**

To determine the first term and the common ratio of a convergent geometric series, additional information is required, such as the sum of the series or specific terms within the series. A geometric series converges if and only if $$ |r| < 1 $$.

**Please provide additional details or information** related to the geometric series (such as the sum, a specific term, or another condition) so that I can assist you in solving parts **3.3.1** and **3.3.2**.

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Feel free to provide the missing details for Question 3.3, and I'll be happy to help further!
**Simplified Answers:** 1. **Question 3.1:** - The 8th term of the sequence is $$ \frac{3}{128} $$. 2. **Question 3.2:** - The numerical value of $$ \frac{x}{y} $$ is **-1**. 3. **Question 3.3:** - **3.3.1** First term: Cannot determine without additional information. - **3.3.2** Common ratio, $$ r $$: Cannot determine without additional information. **Note:** For Question 3.3, more details are needed to find the first term and the common ratio of the convergent geometric series.

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