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 the given series, we first need to identify the patterns in both series. For the first series \( 11 + 8 + 5 + 2 + \ldots - 166 \): This series is an arithmetic series with the first term \( a = 11 \) and a common difference \( d = -3 \). Let's find the number of terms in the series. The \( n \)-th term \( a_n \) can be represented as: \[ a_n = a + (n-1) \cdot d = 11 + (n-1)(-3) = 11 - 3(n-1). \] To find the highest term, we set \( a_n = -166 \): \[ -166 = 11 - 3(n-1) \\ -166 - 11 = -3(n-1) \\ -177 = -3(n-1) \\ n-1 = 59 \\ n = 60. \] Thus, there are 60 terms in this series. The sum \( S_n \) of an arithmetic series can be calculated using: \[ S_n = \frac{n}{2} (a + a_n) = \frac{60}{2} (11 - 166) = 30(-155) = -4650. \] Now for the second series \( -12 - 5 + 2 + 9 + \ldots + 121 \): This series is also an arithmetic series where the first term \( a = -12 \) and the common difference \( d = 7 \). To find the number of terms, we’ll find when the \( n \)-th term equals 121: \[ a_n = -12 + (n-1) \cdot 7 = 121 \\ 121 + 12 = 7(n-1) \\ 133 = 7(n-1) \\ n-1 = 19 \\ n = 20. \] So, there are 20 terms in this series. Now calculating the sum \( S_n \): \[ S_n = \frac{n}{2}(a + a_n) = \frac{20}{2}(-12 + 121) = 10(109) = 1090. \] Finally, summarizing both results: - The sum of the first series \( 1.1 \) is \( -4650 \). - The sum of the second series \( 1.3 \) is \( 1090 \).