2 Determine the sum of the first 40 terms of the arithmetic series \( 2.1 \quad 2+10+18+\ldots \) \( 2.2 16+10+4-2-\ldots \) \( 2.3 1+2.5+4+\ldots \) \( n=40 \)

To find the sum of the first 40 terms of each arithmetic series, we can use the formula for the sum of the first \( n \) terms of an arithmetic series: \[ S_n = \frac{n}{2} \times (a + l) \] where \( S_n \) is the sum, \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term (which can be found using \( l = a + (n-1) \cdot d \), where \( d \) is the common difference). ### 2.1 \( 2 + 10 + 18 + \ldots \) - First term \( a = 2 \) - Common difference \( d = 10 - 2 = 8 \) - 40th term \( l = 2 + (40 - 1) \cdot 8 = 2 + 312 = 314 \) Now plug into the formula: \[ S_{40} = \frac{40}{2} \times (2 + 314) = 20 \times 316 = 6320 \] ### 2.2 \( 16 + 10 + 4 - 2 - \ldots \) - First term \( a = 16 \) - Common difference \( d = 10 - 16 = -6 \) - 40th term \( l = 16 + (40 - 1) \cdot (-6) = 16 - 234 = -218 \) Now plug into the formula: \[ S_{40} = \frac{40}{2} \times (16 + (-218)) = 20 \times (-202) = -4040 \] ### 2.3 \( 1 + 2.5 + 4 + \ldots \) - First term \( a = 1 \) - Common difference \( d = 2.5 - 1 = 1.5 \) - 40th term \( l = 1 + (40 - 1) \cdot 1.5 = 1 + 58.5 = 59.5 \) Now plug into the formula: \[ S_{40} = \frac{40}{2} \times (1 + 59.5) = 20 \times 60.5 = 1210 \] In conclusion, the sums of the first 40 terms are: - \( 2.1: 6320 \) - \( 2.2: -4040 \) - \( 2.3: 1210 \)