\begin{tabular}{l}\hline ACTIVITIESIASSESSMENTS \\ \hline - Classwork/Homework \\ 1. Consider the sequence: \( 12 ; 4 ; 14 ; 7 ; 18 ; 10 ; \ldots \) \\ 1.1 Write down the next TWO terms if the given pattern continues. \\ 1.2 Calculate the value of the \( 50^{\text {th }} \) term of the sequence. \\ 1.3 Write down the value of \( 131^{\text {th }} \) term of the sequence. \\ 2. The following sequence is a combination of an arithmetic and a geometric sequence: \\ \( 3 ; 3 ; 9 ; 6 ; 15 ; 12 ; \ldots \) \\ 2.1 Write down the next TWO terms. \\ 2.2 Calculate \( T_{22}-T_{21} \). \\ 2.3 Prove that ALL the terms of this infinite sequence will be divisible by 3 . \end{tabular}

Sure, let's work through each of the problems step by step. --- ## **1. Sequence Analysis** Given Sequence: **12 ; 4 ; 14 ; 7 ; 18 ; 10 ; …** ### **1.1 Next TWO Terms** **Pattern Identification:** - **Odd-Positioned Terms (1st, 3rd, 5th, ...):** 12, 14, 18, ... - **Differences:** +2 (14 - 12), +4 (18 - 14), ... - **Pattern:** The difference increases by 2 each time. - **Next Difference:** +6 - **Next Odd Term:** 18 + 6 = **24** - **Even-Positioned Terms (2nd, 4th, 6th, ...):** 4, 7, 10, ... - **Differences:** +3 each time - **Next Even Term:** 10 + 3 = **13** **Next TWO Terms:** **24** and **13** --- ### **1.2 Value of the \( 50^{\text{th}} \) Term** **Determining the Position:** - **50 is Even**, so it follows the even-positioned pattern. **Even-Term Formula:** \[ a_n = 3k + 1 \] where \( k = \frac{n}{2} \) \[ k = \frac{50}{2} = 25 \] \[ a_{50} = 3 \times 25 + 1 = 75 + 1 = 76 \] **\( 50^{\text{th}} \) Term:** **76** --- ### **1.3 Value of the \( 131^{\text{th}} \) Term** **Determining the Position:** - **131 is Odd**, so it follows the odd-positioned pattern. **Odd-Term Formula:** \[ a_n = \frac{n^2 + 47}{4} \] \[ a_{131} = \frac{131^2 + 47}{4} \] \[ = \frac{17161 + 47}{4} \] \[ = \frac{17208}{4} \] \[ = 4302 \] **\( 131^{\text{th}} \) Term:** **4302** --- ## **2. Combined Arithmetic and Geometric Sequence** Given Sequence: **3 ; 3 ; 9 ; 6 ; 15 ; 12 ; …** ### **2.1 Next TWO Terms** **Pattern Identification:** - **Odd-Positioned Terms (1st, 3rd, 5th, ...):** 3, 9, 15, ... - **Differences:** +6 each time - **Next Odd Term:** 15 + 6 = **21** - **Even-Positioned Terms (2nd, 4th, 6th, ...):** 3, 6, 12, ... - **Pattern:** Each term is multiplied by 2 - **Next Even Term:** 12 × 2 = **24** **Next TWO Terms:** **21** and **24** --- ### **2.2 Calculate \( T_{22} - T_{21} \)** **Determining the Terms:** - **\( T_{21} \) (Odd Position):** \[ a_k = 3 + 6(k - 1) \] \[ k = \frac{21 + 1}{2} = 11 \] \[ a_{21} = 3 + 6(11 - 1) = 3 + 60 = 63 \] - **\( T_{22} \) (Even Position):** \[ a_k = 3 \times 2^{k-1} \] \[ k = \frac{22}{2} = 11 \] \[ a_{22} = 3 \times 2^{10} = 3 \times 1024 = 3072 \] **Calculation:** \[ T_{22} - T_{21} = 3072 - 63 = 3009 \] **Result:** **3009** --- ### **2.3 Proof that ALL Terms are Divisible by 3** **Odd-Positioned Terms:** - **Formula:** \[ a_k = 3 + 6(k - 1) \] \[ = 3(1 + 2(k - 1)) \] - Clearly, \( a_k \) is a multiple of 3. **Even-Positioned Terms:** - **Formula:** \[ a_k = 3 \times 2^{k-1} \] - Regardless of the value of \( k \), \( a_k \) is \( 3 \) multiplied by an integer \( 2^{k-1} \), making it a multiple of 3. **Conclusion:** All terms, whether odd or even-positioned, are multiples of **3**. --- ## **Summary of Answers** 1. **Sequence 12; 4; 14; 7; 18; 10; …** - **1.1 Next TWO Terms:** 24 and 13 - **1.2 \( 50^{\text{th}} \) Term:** 76 - **1.3 \( 131^{\text{th}} \) Term:** 4302 2. **Sequence 3; 3; 9; 6; 15; 12; …** - **2.1 Next TWO Terms:** 21 and 24 - **2.2 \( T_{22} - T_{21} \):** 3009 - **2.3 Proof of Divisibility by 3:** All terms are divisible by 3. --- If you have any further questions or need additional explanations, feel free to ask!