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 \( \frac{T_{1}+T_{3}^{\prime}+T_{5}+T_{7}+\cdots+T_{499}}{T_{2}+T_{4}^{4}+T_{6}+T_{8}+\cdots+T_{500}} \)

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 **\(-2\)**.