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Darwin Joseph Santos

Mark Joseph Salazar

Cluadine Doma

Maricon Hollon

Randolph Sampang

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- 1. Arithmetic Series and Arithmetic Mean Prepared By: Salazar, Mark Joseph Sampang, Randolph Brian Santos, Darwin Joseph Doma, Cluadine Hollon, Maricon
- 2. Arithmetic Series
- 3. Arithmetic Series A series such as (3 + 7 + 11 + 15 + ··· + 99 or 10 + 20 + 30 + ··· + 1000) which has a constant difference between terms. first term is a1 common difference is d number of terms is n sum of an arithmetic series is Sn An arithmetic series is the sum of an arithmetic sequence.
- 4. Arithmetic Series Formula: or ~when an is given
- 5. Arithmetic Series Example #1: 3 + 7 + 11 + 15 + ··· + 99 has a1 = 3 and d = 4. To find n, use the explicit formula for an arithmetic sequence. We solve: 3 + (n – 1)·4 = 99 to get n = 25.
- 6. Arithmetic Series Example #2: Find the sum of the first 12 positive even integers. positive even integers: 2, 4, 6, 8, ... n = 12; a1 = 2, d = 2 We are missing a12, for the sum formula so we will use = 12/2[2(2) + (12 – 1)2] = 6[4 + 22] = 6(26) = 156
- 7. Arithmetic Series Activity: Find the sum of each arithmetic series. 1. Find the sum of the sequence -8, -5, -2, ..., 7 2. Find the sum of the first 10 positive integers 3. Find the sum of the first 20 terms of the sequence 4, 6, 8, 10, ... Answers: 1. -3 2. 55 3. 460
- 8. Arithmetic Mean
- 9. Arithmetic Mean The numbers between arithmetic extremes are called arithmetic mean, found in an arithmetic sequence wherein each term is obtained by adding a fixed value called the common difference. Example: 4, 7, 10, 13, 16 The arithmetic means are 7, 10 and 13 9, 15, 21 The arithmetic mean is 15
- 10. Let’s Try! ① Insert 3 arithmetic means between 1 and 17 1, _ , _ , _ , 17 a5 = 1 + (5-1)d 17 = 1 + (5-1)d 17 = 1 + 4d 17 - 1 = 4d 16 = 4d d = 4 an = a1 + (n-1)d a2 = a1 + d a2 = 1 + 4 a2 = 5 a3 = a1 + 2d a3 = 1 + (2)4 a3= 9 a4 = a1 + 3d a4= 1 + (3)4 a4= 13 ② Insert arithmetic means between 95 and 185 95, _ , 185 an = a1 + (n-1)d 185 = 95 + (3-1)d 185 = 95 + 2d 185 - 95 = 2d 90 = 2d d = 45 a2 = a1 + d a2 = 95 + 45 a2 = 140
- 11. Word Problem ③ John recruited 2 persons for the networking business. After a week, he recruited 5 persons again and on the 5th week of recruitment, he recruited another 14 persons for the networking business. If this continues, how many persons did John already recruited after the 6th week of recruitment? an = ? a1 = 2 d = 3 n = 6 Sn = n/2 [2a1 + (n - 1)d] S6= 6/2 [2(2) + (6 - 1)3] S6= 3 [4 + (5)3] S6 = 3 [4 + (15)] S6 = 3 [19] S6 = 57 John already recruited 57 persons after 6 weeks of recruitment

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