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

To determine the sum of each of the series, let's analyze both sequentially. 1.1 The series \( 11 + 8 + 5 + 2 + \ldots - 166 \) is an arithmetic series where the first term \( a = 11 \) and the common difference \( d = -3 \). The last term is \(-166\). To find the number of terms \( n \), we can use the nth-term formula of an arithmetic sequence: \[ a_n = a + (n-1) d \] Setting \( a_n = -166 \): \[ -166 = 11 + (n - 1)(-3) \] Solving gives: \[ -166 - 11 = (n - 1)(-3) \] \[ -177 = (n - 1)(-3) \] \[ n - 1 = 59 \quad \Rightarrow \quad n = 60 \] Now, using the sum formula for an arithmetic series: \[ S_n = \frac{n}{2} (a + a_n) \] Substituting the values: \[ S_{60} = \frac{60}{2} (11 - 166) = 30 \times (-155) = -4650 \] 1.3 The series \(-12 - 5 + 2 + 9 + \ldots + 121\) is another arithmetic series where the first term \( a = -12 \) and the common difference \( d = 7 \). To find the number of terms \( n \), we use the last term: Setting \( a_n = 121 \): \[ 121 = -12 + (n - 1)(7) \] Solving gives: \[ 121 + 12 = (n - 1)(7) \] \[ 133 = (n - 1)(7) \] \[ n - 1 = \frac{133}{7} = 19 \quad \Rightarrow \quad n = 20 \] Now, using the sum formula: \[ S_n = \frac{n}{2} (a + a_n) \] Substituting the values: \[ S_{20} = \frac{20}{2} (-12 + 121) = 10 \times 109 = 1090 \] So, the sums of the series are: 1.1. -4650 1.3. 1090