\[ \mathrm{T}_{n}=a+(n-1) d=12+(n-1)(5)=5 n+7 \] EXERCISE 5 Determine the sum of the series: \( 1.1 \quad 11+8+5+2+\ldots-166 \). \( 1.21+5+9+13+\ldots+113 \) \( 1.3-12-5+2+9+\ldots+121 \) Determine the sum of the first 40 terms of the arithmetic series \( 2.1 \quad 2+10+18+\ldots \quad n=40 \) \( 2.216+10+4-2-\ldots \)

To find the sum of the series, let's break it down into each specific sequence. 1. **For the first series (11 + 8 + 5 + 2 + ... - 166)**: This is a decreasing arithmetic sequence where the first term \( a = 11 \) and the common difference \( d = -3 \). To find the number of terms until reaching -166, use the formula \( T_n = a + (n-1)d \): \[ -166 = 11 + (n-1)(-3) \implies -166 = 11 - 3(n-1) \implies -177 = -3(n-1) \implies n-1 = 59 \implies n = 60 \] The sum of the first \( n \) terms can be calculated using the sum formula \( S_n = \frac{n}{2} (a + T_n) \): \[ S_{60} = \frac{60}{2} (11 + (-166)) = 30 \times (-155) = -4650 \] 2. **For the second series (1 + 5 + 9 + 13 + ... + 113)**: This is an increasing arithmetic series with \( a = 1 \) and \( d = 4 \). To find the number of terms until 113: \[ 113 = 1 + (n-1)4 \implies 112 = 4(n-1) \implies n-1 = 28 \implies n = 29 \] The sum of the first 29 terms: \[ S_{29} = \frac{29}{2} (1 + 113) = \frac{29}{2} \times 114 = 29 \times 57 = 1653 \] 3. **For the third series (-12 - 5 + 2 + 9 + ... + 121)**: Here, \( a = -12 \) and \( d = 7 \). Finding the number of terms to 121: \[ 121 = -12 + (n-1)7 \implies 133 = 7(n-1) \implies n-1 = 19 \implies n = 20 \] Therefore the sum of the first 20 terms: \[ S_{20} = \frac{20}{2} (-12 + 121) = 10 \times 109 = 1090 \] 4. **For the arithmetic series (2 + 10 + 18 + ... with n = 40)**: Here, \( a = 2 \) and \( d = 8 \). The sum of the first 40 terms is: \[ S_{40} = \frac{40}{2} (2 + T_{40}) \] First, we need to find \( T_{40} \): \[ T_{40} = 2 + (40-1)8 = 2 + 312 = 314 \] Now using the sum formula: \[ S_{40} = 20(2 + 314) = 20 \times 316 = 6320 \] 5. **For the series (16 + 10 + 4 - 2 - ...)**: Here, \( a = 16 \) and \( d = -6 \). To find the number of terms until reaching the stopping point: Utilizing \( T_n = a + (n-1)d \) to see where it approaches a stop or a pattern that can reasonably sum. Since no terminating term is set as per your given data, you may prefer summing a first set number (be it 40) accordingly to the derived pattern. If you need clarification, adjustments, or further breakdowns, let me know!