ACTIVITY #3: ARITHMETIC SERIES A. Find the sum of the following arithmetic series. 1. 4, 11, (16th terms) www 2. 19, 13, ... (10th terms) 3.-9,-1, (8th terms) B. Answer the following and show your solutions. 1. Find the sum of the first 80 natural numbers. 2. Find the sum of the multiples of 7 from 1 to 100. 3. Find the sum of the multiples of 9 from 1 to 300.​

Answer:A.1. Find the sum of the arithmetic series 4, 11, ... (16th term) First, we need to find the common difference (d) of the arithmetic series: d = 11 - 4 = 7. The formula for the nth term of an arithmetic series is a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, n is the number of terms, and d is the common difference. The 16th term is: a_{16} = 4 + (16-1)7 = 4 + 105 = 109. The formula for the sum of an arithmetic series is S_n = \frac{n}{2}(a_1 + a_n), where S_n is the sum of the first n terms. Therefore, the sum of the first 16 terms is: S_{16} = \frac{16}{2}(4 + 109) = 8(113) = 904. Sum: \boxed{904}   A.2. Find the sum of the arithmetic series 19, 13, ... (10th term) The common difference is d = 13 - 19 = -6. The 10th term is: a_{10} = 19 + (10-1)(-6) = 19 - 54 = -35. The sum of the first 10 terms is: S_{10} = \frac{10}{2}(19 + (-35)) = 5(-16) = -80. Sum: \boxed{-80}   A.3. Find the sum of the arithmetic series -9, -1, ... (8th term) The common difference is d = -1 - (-9) = 8. The 8th term is: a_8 = -9 + (8-1)8 = -9 + 56 = 47. The sum of the first 8 terms is: S_8 = \frac{8}{2}(-9 + 47) = 4(38) = 152. Sum: \boxed{152}   B.1. Find the sum of the first 80 natural numbers. This is an arithmetic series with a_1 = 1, a_{80} = 80, and n = 80. The sum is: S_{80} = \frac{80}{2}(1 + 80) = 40(81) = 3240. Sum: \boxed{3240}   B.2. Find the sum of the multiples of 7 from 1 to 100. The multiples of 7 are 7, 14, 21, ..., 98. This is an arithmetic series with a_1 = 7, d = 7, and we need to find the number of terms. a_n = a_1 + (n-1)d, so 98 = 7 + (n-1)7. Solving for n, we get n = 14. The sum is: S_{14} = \frac{14}{2}(7 + 98) = 7(105) = 735. Sum: \boxed{735}   B.3. Find the sum of the multiples of 9 from 1 to 300. The multiples of 9 are 9, 18, 27, ..., 297. This is an arithmetic series with a_1 = 9, d = 9. We need to find the number of terms. a_n = a_1 + (n-1)d, so 297 = 9 + (n-1)9. Solving for n, we get n = 33. The sum is: S_{33} = \frac{33}{2}(9 + 297) = \frac{33}{2}(306) = 33(153) = 5049. Sum: \boxed{5049}

Answered by thegrilledbacon | 2025-07-02