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# Arithmetic Series

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# Arithmetic Series

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### Arithmetic Series

1. 1. Grade 10 – Mathematics Quarter I ARITHMETIC SERIES
2. 2. Objectives: •define arithmetic series; and •find the sum of n terms of arithmetic sequence given the first few terms of an arithmetic sequence.
3. 3. The sum of the terms of an arithmetic sequence forms an arithmetic series. The sum of the first n terms of a sequence, called a partial sum, is denoted by 𝑆 𝑛.
4. 4. The sum of n a terms of an arithmetic sequence is given by: 𝑖𝑓 𝑎 𝑛 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛, 𝑆 𝑛 = 𝑛 2 𝑎1 + 𝑎 𝑛 𝑖𝑓 𝑎 𝑛 𝑖𝑠 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛, 𝑆 𝑛 = 𝑛 2 2𝑎 + 𝑛 − 1 𝑑
5. 5. Find the sum of the first 12 terms of the arithmetic sequence 50, 47, 44, 41, 38, … Solution: 𝑎 = 50, 𝑑 = −3, 𝑛 = 12 𝑆 𝑛 = 𝑛 2 2𝑎 + 𝑛 − 1 𝑑 𝑆 𝑛 = 12 2 2 50 + 12 − 1 − 3 𝑆12= 6 100 − 33 𝑺 𝟏𝟐= 402
6. 6. Find the sum of the first 18 terms of the arithmetic sequence 3, 5, 7, 9, 11, … Solution: 𝑎 = 3, 𝑑 = 2, 𝑛 = 18 𝑆 𝑛 = 𝑛 2 2𝑎 + 𝑛 − 1 𝑑 𝑆 𝑛 = 18 2 2 3 + 18 − 1 2 𝑆18= 9 6 + 34 𝑺 𝟏𝟖= 40
7. 7. Find the sum of the first 30 natural numbers. Solution: 𝑎1 = 1, 𝑎 𝑛 = 30, 𝑛 = 30 𝑆 𝑛 = 𝑛 2 𝑎1 + 𝑎 𝑛 𝑆 𝑛 = 30 2 1 + 30 𝑆30= 15 31 𝑺 𝟑𝟎= 465
8. 8. Find the sum of the first 50 multiples of 5. Solution: 𝑎1 = 5, 𝑎 𝑛 = 50, 𝑛 = 50 𝑆 𝑛 = 𝑛 2 𝑎1 + 𝑎 𝑛 𝑆 𝑛 = 50 2 5 + 50 𝑆50= 25 55 𝑺 𝟓𝟎= 1375
9. 9. Find the sum of all multiples of 3 between 1 and 100. Solution: 𝑎1 = 3, 𝑎 𝑛 = 99, 𝑑 = 3 First, we determine how many multiples of 3 there are between 1 and 100. 𝑎 𝑛 = 𝑎 + 𝑛 − 1 𝑑 99 = 3 + 𝑛 − 1 3 99 = 3 + 3𝑛 − 3 99 = 3𝑛 𝒏 = 𝟑𝟑 𝑆 𝑛 = 𝑛 2 𝑎1 + 𝑎 𝑛 𝑆33 = 33 2 3 + 99 𝑺 𝟑𝟑 = 1, 683
10. 10. Find the sum of all multiples of 6 between 1 and 100. Solution: 𝑎1 = 6, 𝑎 𝑛 = 96, 𝑑 = 6 First, we determine how many multiples of 6 there are between 1 and 100. 𝑎 𝑛 = 6 + 𝑛 − 1 𝑑 96 = 6 + 𝑛 − 1 6 96 = 6 + 6𝑛 − 6 96 = 6𝑛 𝒏 = 𝟏𝟔 𝑆 𝑛 = 𝑛 2 𝑎1 + 𝑎 𝑛 𝑆33 = 16 2 6 + 96 𝑺 𝟑𝟑 = 𝟖𝟏𝟔