ACTIVITY 91. Find the sum of the first 17 terms of the arithmeticsequence 2, 5, 8,...2. Find the sum of the first 25 terms of the arithmeticsequence 20, 18, 16, ...3. Find the sum of the first 20 terms of an arithmeticsequence whose a₁ = 25 and a20 = -13.4. Find the sum of the first 200 natural numbers.5. Find the sum of all multiples of 6 between 1 and 100.
Answer (1)
Answer:1. **Sum of the first 17 terms of the sequence 2, 5, 8,...** - The first term \( a_1 = 2 \) and the common difference \( d = 3 \). - The 17th term \( a_{17} = a_1 + (17 - 1)d = 2 + 16 \times 3 = 50 \). - Sum \( S_{17} = \frac{17}{2} \times (a_1 + a_{17}) = \frac{17}{2} \times (2 + 50) = 17 \times 26 = 442 \).2. **Sum of the first 25 terms of the sequence 20, 18, 16,...** - The first term \( a_1 = 20 \) and the common difference \( d = -2 \). - The 25th term \( a_{25} = a_1 + (25 - 1)d = 20 + 24 \times (-2) = -28 \). - Sum \( S_{25} = \frac{25}{2} \times (a_1 + a_{25}) = \frac{25}{2} \times (20 - 28) = 25 \times (-4) = -100 \).3. **Sum of the first 20 terms of a sequence with \( a_1 = 25 \) and \( a_{20} = -13 \)** - The 20th term \( a_{20} = a_1 + 19d = -13 \), so \( 25 + 19d = -13 \) gives \( d = -2.105 \). - Sum \( S_{20} = \frac{20}{2} \times (a_1 + a_{20}) = 10 \times (25 - 13) = 10 \times 12 = 120 \).4. **Sum of the first 200 natural numbers** - Use the formula \( S_n = \frac{n(n + 1)}{2} \). - Sum \( S_{200} = \frac{200 \times 201}{2} = 20100 \).5. **Sum of all multiples of 6 between 1 and 100** - The multiples of 6 are 6, 12, 18, ..., 96. - This is an arithmetic sequence with \( a_1 = 6 \), \( d = 6 \), and \( a_n = 96 \). - The number of terms \( n \) is \( \frac{96 - 6}{6} + 1 = 16 \). - Sum \( S_{16} = \frac{16}{2} \times (6 + 96) = 8 \times 102 = 816 \).
