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Understanding Arithmetic Sequences and Series: A Comprehensive Guide

This resource by Jeff Bivin from Lake Zurich High School provides a detailed exploration of arithmetic sequences and series. It covers finding the nth term, working with first terms and common differences, and calculating sums of sequences. Through examples and step-by-step explanations, the guide helps students grasp key concepts in arithmetic sequences including formula derivations and applications like sums of integers and multiples. Ideal for high school students seeking to enhance their understanding of this mathematical topic.

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Understanding Arithmetic Sequences and Series: A Comprehensive Guide

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  1. Arithmetic Sequences & Series By: Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: April 28, 2006

  2. Arithmetic Sequences 5, 8, 11, 14, 17, 20, … 3n+2, … -4, 1, 6, 11, 16, … 5n – 9, . . . 11, 7, 3, -1, -5, … -4n + 15, . . . Jeff Bivin -- LZHS

  3. nth term of arithmetic sequence an = a1 + d(n – 1) Jeff Bivin -- LZHS

  4. Find the nth term of an arithmetic sequence First term is 8 Common difference is 3 an = a1 + d(n – 1) an = 8 + 3(n – 1) an = 8 + 3n – 3 an = 3n + 5 Jeff Bivin -- LZHS

  5. Finding the nth term First term is -6 common difference is 7 an = a1 + d(n – 1) an = -6 + 7(n – 1) an = -6 + 7n – 7 an = 7n - 13 Jeff Bivin -- LZHS

  6. Finding the nth term First term is 23 common difference is -4 an = a1 + d(n – 1) an = 23 + -4(n – 1) an = 23 -4n +4 an = -4n + 27 Jeff Bivin -- LZHS

  7. Finding the 100th term a1 = 5 d = 6 n = 100 5, 11, 17, 23, 29, . . . an = a1 + d(n – 1) a100 = 5 + 6(100 – 1) a100 = 5 + 6(99) a100 = 5 + 594 a100 = 599 Jeff Bivin -- LZHS

  8. Finding the 956th term a1 = 156 d = -16 n = 956 156, 140, 124, 108, . . . an = a1 + d(n – 1) a956 = 156 + -16(956 – 1) a956 = 156 - 16(955) a956 = 156 - 15280 a956 = -15124 Jeff Bivin -- LZHS

  9. Find the Sum of the integers from 1 to 100 S100 = 1 + 2 + 3 +…+ 49 + 50 + 51 + 52 +…+ 98 + 99 + 100 S100 = 100 + 99 + 98 +…+ 52+51 + 50 + 49 +…+ 3 + 2 + 1 2S100 = 101+101+101+…+101+101+101+101+…+101+101+101 2S100 = 100 (101) Jeff Bivin -- LZHS

  10. Summing it up Sn = a1 + (a1 + d) + (a1 + 2d) + …+ an Sn = an + (an - d) + (an - 2d) + …+ a1 Jeff Bivin -- LZHS

  11. 1 + 4 + 7 + 10 + 13 + 16 + 19 a1 = 1 an = 19 n = 7 Jeff Bivin -- LZHS

  12. 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 a1 = 4 an = 24 n = 11 Jeff Bivin -- LZHS

  13. Find the sum of the integers from 1 to 100 a1 = 1 an = 100 n = 100 Jeff Bivin -- LZHS

  14. Find the sum of the multiples of 3 between 9 and 1344 Sn = 9 + 12 + 15 + . . . + 1344 a1 = 9 an = 1344 d = 3 Jeff Bivin -- LZHS

  15. Find the sum of the multiples of 7 between 25 and 989 Sn = 28 + 35 + 42 + . . . + 987 a1 = 28 an = 987 d = 7 Jeff Bivin -- LZHS

  16. Evaluate Sn = 16 + 19 + 22 + . . . + 82 a1 = 16 an = 82 d = 3 n = 23 Jeff Bivin -- LZHS

  17. Evaluate Sn = -29 - 31 - 33 + . . . - 199 a1 = -29 an = -199 d = -2 n = 86 Jeff Bivin -- LZHS

  18. Find the sum of the multiples of 11 that are 4 digits in length Sn = 10 01+ 1012 + 1023 + ... + 9999 a1 = 1001 an = 9999 d = 11 Jeff Bivin -- LZHS

  19. Review -- Arithmetic Sum of n terms nth term Jeff Bivin -- LZHS

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