The following sequence is a combination of arithmetic and geometric sequence: \( \begin{array}{llll}3 ; 3 ; 9 ; 6 ; 15 ; 12 ; \ldots \\ \text { 4.1.1 Write down the next TWO } & \text { (2) L1 } & \text { 4.1.2 Determine } T_{20}-T_{21}\end{array} \)

Let's analyze the given sequence and solve the problems step by step. **Given Sequence:** \[ 3; \, 3; \, 9; \, 6; \, 15; \, 12; \, \ldots \] The sequence alternates between two different patterns: - **Odd-positioned terms (1st, 3rd, 5th, ...):** These terms form an **arithmetic sequence**. - **Even-positioned terms (2nd, 4th, 6th, ...):** These terms form a **geometric sequence**. --- ### **4.1.1 Write Down the Next Two Terms** **1. Identify the Patterns:** - **Odd Positions (Arithmetic Sequence):** - **Terms:** 3, 9, 15, ... - **Common Difference (d):** \( 9 - 3 = 6 \) - **General Formula:** \( T_n = 3 + (k-1) \times 6 = 6k - 3 \) where \( k \) is the position in the odd sequence. - **Even Positions (Geometric Sequence):** - **Terms:** 3, 6, 12, ... - **Common Ratio (r):** \( \frac{6}{3} = 2 \) - **General Formula:** \( T_n = 3 \times 2^{k-1} \) where \( k \) is the position in the even sequence. **2. Determine the Next Two Terms:** - **7th Term (Odd Position):** \[ T_7 = 3 + 3 \times 6 = 21 \] - **8th Term (Even Position):** \[ T_8 = 3 \times 2^{4-1} = 3 \times 8 = 24 \] ****Answer:** The next two terms are **21** and **24**. --- ### **4.1.2 Determine \( T_{20} - T_{21} \)** **1. Find \( T_{20} \) and \( T_{21} \):** - **20th Term (\( T_{20} \))** \( \) is an **even-positioned term**: \[ T_{20} = 3 \times 2^{\frac{20}{2} - 1} = 3 \times 2^{9} = 3 \times 512 = 1536 \] - **21st Term (\( T_{21} \))** is an **odd-positioned term**: \[ T_{21} = 3 \times 21 = 63 \] **2. Calculate the Difference:** \[ T_{20} - T_{21} = 1536 - 63 = 1473 \] ****Answer:** \( T_{20} - T_{21} = 1473 \) --- **Summary:** - **Next Two Terms:** 21 and 24 - **Difference \( T_{20} - T_{21} \):** 1473