1 Determine the sum of the series: \( 1.1 \quad 11+8+5+2+\ldots-166 \). \( 1.3 \quad-12-5+2+9+\ldots+121 \)

Let's solve each part of the problem step by step. ### 1.1. Sum of the Series: \( 11 + 8 + 5 + 2 + \ldots -166 \) **Step 1: Identify the Type of Series** This is an **arithmetic series** where each term decreases by a common difference. **Step 2: Determine the Common Difference (\( d \))** \[ d = 8 - 11 = -3 \] **Step 3: Find the Number of Terms (\( n \))** Use the formula for the \( n \)-th term of an arithmetic series: \[ a_n = a_1 + (n-1)d \] Given \( a_n = -166 \) and \( a_1 = 11 \): \[ -166 = 11 + (n-1)(-3) \\ -166 = 11 - 3n + 3 \\ -166 = 14 - 3n \\ -180 = -3n \\ n = 60 \] **Step 4: Calculate the Sum (\( S_n \))** Use the sum formula for an arithmetic series: \[ S_n = \frac{n}{2} \times (a_1 + a_n) \\ S_{60} = \frac{60}{2} \times (11 + (-166)) \\ S_{60} = 30 \times (-155) \\ S_{60} = -4650 \] **Answer for 1.1:** The sum of the series is **-4,650**. --- ### 1.3. Sum of the Series: \( -12 -5 +2 +9 + \ldots +121 \) **Step 1: Identify the Type of Series** This is an **arithmetic series** where each term increases by a common difference. **Step 2: Determine the Common Difference (\( d \))** \[ d = -5 - (-12) = 7 \] **Step 3: Find the Number of Terms (\( n \))** Use the formula for the \( n \)-th term of an arithmetic series: \[ a_n = a_1 + (n-1)d \] Given \( a_n = 121 \) and \( a_1 = -12 \): \[ 121 = -12 + (n-1) \times 7 \\ 121 + 12 = 7(n-1) \\ 133 = 7(n-1) \\ n-1 = 19 \\ n = 20 \] **Step 4: Calculate the Sum (\( S_n \))** Use the sum formula for an arithmetic series: \[ S_n = \frac{n}{2} \times (a_1 + a_n) \\ S_{20} = \frac{20}{2} \times (-12 + 121) \\ S_{20} = 10 \times 109 \\ S_{20} = 1090 \] **Answer for 1.3:** The sum of the series is **1,090**. ## Final Summary 1. **1.1:** The sum of \( 11 + 8 + 5 + 2 + \ldots -166 \) is **-4,650**. 2. **1.3:** The sum of \( -12 -5 +2 +9 + \ldots +121 \) is **1,090**.