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

To calculate the sum of the series in \( 1.1 \) and \( 1.3 \), we first analyze each series separately and identify patterns. ### Series \( 1.1: 11 + 8 + 5 + 2 + \ldots - 166 \) 1. **Identify the pattern**: The series starts at 11 and decreases by 3 each time: - First term: \( a_1 = 11 \) - Second term: \( a_2 = 8 \) - Third term: \( a_3 = 5 \) - Fourth term: \( a_4 = 2 \) - Continuing this pattern, the nth term can be expressed as: \[ a_n = 11 - 3(n-1) = 14 - 3n \] 2. **Determine when the series equals -166**: Set \( a_n = -166 \): \[ 14 - 3n = -166 \] \[ 3n = 180 \implies n = 60 \] 3. **Sum the terms**: This series has 60 terms. - Formula for the sum of an arithmetic series \( S_n = \frac{n}{2} (a_1 + a_n) \): - Here \( n = 60 \), \( a_1 = 11 \), and \( a_n = -166 \): \[ S_{60} = \frac{60}{2} (11 - 166) = 30 \cdot (-155) = -4650 \] ### Series \( 1.3: -12 - 5 + 2 + 9 + \ldots + 121 \) 1. **Identify the pattern**: The series starts from -12 and increases by 7 for the odd terms and then decreases by 5 for the even terms: - First term: \( a_1 = -12 \) - Second term: \( a_2 = -5 \) - Third term: \( a_3 = 2 \) - Fourth term: \( a_4 = 9 \) - Fifth term: \( a_5 = 16 \) - The nth term pattern can be seen as: \[ \text{Odd indices: } 2 + 7(n-2) \quad \text{ and } \quad \text{Even indices: } -12 + 7(n-1) \] 2. **Determine when the sum reaches 121**: We need to calculate how many terms up to 121: - The maximum term \( a_n \) occurs at an even term since it’s the last term at 121: Set \( a_n = 121 \): \[ -12 + 7m = 121 \implies 7m = 133 \implies m = 19 \] so, there are \( n = 20 \) terms total. 3. **Sum the terms**: Use the sum formula where \( S_n = \frac{n}{2} (a_1 + a_n) \). - Here \( a_n = 121 \), \( n = 20 \), and \( a_1 = -12 \): \[ S_{20} = \frac{20}{2} (-12 + 121) = 10 \cdot 109 = 1090 \] ### Final Result The sums of the series are: - For \( 1.1 \): \(-4650\) - For \( 1.3 \): \(1090\)