Given a geometric sequence. $$ 36 ;-18 ; 9 ; \ldots $$ 12.1 Determine the value of $$ r $$, the common ratio. 12.2 Calculate $$ n $$ if $$ T_{2}=\frac{9}{4096} $$. 12.3 Calculate $$ S_{\infty} $$ 12.3 Calculate the value of $$ \frac{T_{1}+T_{3}^{\prime}+T_{5}+T_{7}+\cdots+T_{499}}{T_{2}+T_{4}^{4}+T_{6}+T_{8}+\cdots+T_{500}} $$

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Let's analyze each part of the problem step by step.

### **12.1 Determine the value of $$ r $$, the common ratio.**

A geometric sequence is defined by the relation:
$$
T_{n} = T_{1} \times r^{n-1}
$$
Given the sequence: $$ 36, -18, 9, \ldots $$

- **First Term ($$ T_{1} $$)**: 36
- **Second Term ($$ T_{2} $$)**: -18

The common ratio $$ r $$ is calculated by dividing the second term by the first term:
$$
r = \frac{T_{2}}{T_{1}} = \frac{-18}{36} = -\frac{1}{2}
$$

**Answer:**  
**12.1** The common ratio $$ r $$ is $$-\frac{1}{2}$$.

---

### **12.2 Calculate $$ n $$ if $$ T_{n} = \frac{9}{4096} $$.**

Using the general formula for the $$ n $$-th term of a geometric sequence:
$$
T_{n} = T_{1} \times r^{n-1}
$$
Given:
$$
36 \times \left(-\frac{1}{2}\right)^{n-1} = \frac{9}{4096}
$$

1. **Divide both sides by 36:**
$$
\left(-\frac{1}{2}\right)^{n-1} = \frac{9}{4096 \times 36} = \frac{1}{16384} = \frac{1}{2^{14}}
$$

2. **Since $$\left(-\frac{1}{2}\right)^{n-1} = \frac{1}{2^{14}}$$, the exponent must satisfy:**
$$
2^{-(n-1)} = 2^{-14} \implies n-1 = 14 \implies n = 15
$$

**Answer:**  
**12.2** The value of $$ n $$ is **15**.

---

### **12.3 Calculate $$ S_{\infty} $$**

The sum to infinity $$ S_{\infty} $$ of a geometric series is given by:
$$
S_{\infty} = \frac{T_{1}}{1 - r}
$$
This formula is valid only if $$ |r| < 1 $$.

Given:
$$
r = -\frac{1}{2} \quad \text{and} \quad T_{1} = 36
$$
$$
S_{\infty} = \frac{36}{1 - (-\frac{1}{2})} = \frac{36}{1 + \frac{1}{2}} = \frac{36}{\frac{3}{2}} = 24
$$

**Answer:**  
**12.3** The sum to infinity $$ S_{\infty} $$ is **24**.

---

### **12.4 Calculate the value of $$ \frac{T_{1} + T_{3} + T_{5} + \cdots + T_{499}}{T_{2} + T_{4} + T_{6} + \cdots + T_{500}} $$**

Let's denote:
- **Odd Terms Sum ($$ S_{\text{odd}} $$)**: $$ T_{1} + T_{3} + T_{5} + \cdots + T_{499} $$
- **Even Terms Sum ($$ S_{\text{even}} $$)**: $$ T_{2} + T_{4} + T_{6} + \cdots + T_{500} $$

Given that $$ r = -\frac{1}{2} $$ and $$ T_{1} = 36 $$:
- **Odd Terms:** These form their own geometric series with first term $$ 36 $$ and common ratio $$ \left(-\frac{1}{2}\right)^2 = \frac{1}{4} $$.
- **Even Terms:** These form another geometric series with first term $$ -18 $$ (since $$ T_{2} = 36 \times -\frac{1}{2} $$) and the same common ratio $$ \frac{1}{4} $$.

1. **Number of Terms:**
    - Both $$ S_{\text{odd}} $$ and $$ S_{\text{even}} $$ have 250 terms each.

2. **Sum of Odd Terms:**
    $$
    S_{\text{odd}} = 36 \times \frac{1 - \left(\frac{1}{4}\right)^{250}}{1 - \frac{1}{4}} = 48 \left(1 - \frac{1}{4^{250}}\right)
    $$

3. **Sum of Even Terms:**
    $$
    S_{\text{even}} = -18 \times \frac{1 - \left(\frac{1}{4}\right)^{250}}{1 - \frac{1}{4}} = -24 \left(1 - \frac{1}{4^{250}}\right)
    $$

4. **Ratio:**
    $$
    \frac{S_{\text{odd}}}{S_{\text{even}}} = \frac{48\left(1 - \frac{1}{4^{250}}\right)}{-24\left(1 - \frac{1}{4^{250}}\right)} = -2
    $$

**Answer:**  
**12.4** The value of the given ratio is **$$-2$$**.
**12.1** The common ratio $$ r $$ is $$-\frac{1}{2}$$. **12.2** The value of $$ n $$ is **15**. **12.3** The sum to infinity $$ S_{\infty} $$ is **24**. **12.4** The value of the ratio is **$$-2$$**.

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