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ALGEBRA II HONORS @

ALGEBRA II HONORS @. SECTION 10-2 : ANALYZE ARITHMETIC SEQUENCES and SERIES. ARITHMETIC SEQUENCE : A progression in which one term equals a constant (common difference) added to the preceding term. Write a rule for the nth term of the arithmetic sequence. Then, find a 20 . 1) 3, 5, 7, 9, ….

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ALGEBRA II HONORS @

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  1. ALGEBRA II HONORS @ SECTION 10-2 : ANALYZE ARITHMETIC SEQUENCES and SERIES

  2. ARITHMETIC SEQUENCE : A progression in which one term equals a constant (common difference) added to the preceding term. Write a rule for the nth term of the arithmetic sequence. Then, find a20. 1) 3, 5, 7, 9, … an = 2n + 1, 41 2) 7, 17, 27, 37, … an = 10n – 3, 197

  3. Given the arithmetic sequence : a1a2a3a4a5a6 3 10 17 24 31 38 … NOTE : a1means 1st term, a2the 2nd term, and so on. 3) What is the common difference? SOLUTION : 7 a1= 3 + 7(0) a4= 3 + 7(3) a2= 3 + 7(1) a5= 3 + 7(4) a3= 3 + 7(2) a6= 3 + 7(5) 4) an= ___________________ SOLUTION : an= a1+ (n – 1)d

  4. ARITHMETIC SEQUENCE FORMULA an= a1+ (n – 1)d GREENmeans the formula is to be written on your formulas page. anstands for the nth term a1stands for the first term n is the term number d is the common difference

  5. 5) Write a rule for the nth term of the arithmetic sequence. Then, graph the first six terms of the sequence. a8 = 11, d = 3

  6. 6) Write a rule for the nth term of the arithmetic sequence that has the two given terms. a10 = 47 and a17 = 82 Solution : Write a system of equations using an = a1 + (n – 1)d a10 = a1 + (10 – 1)d -----------> 47 = a1 + 9d a17 = a1 + (17 – 1)d -----------> 82 = a1 + 16d Subtracting, you get -35 = -7d, so d = 5. Substituting in, you get a1 = 2 Therefore, an = 2 + (n – 1)(5)

  7. SERIES :The sum of an arithmetic sequence. Put your writing instruments down and watch Find S100 for the arithmetic series 7 + 13 + 19 + 25 + 31 + … Let’s find a100. an = a1+ (n – 1)d a100= 7 + (100 – 1)(6) = 601 Therefore, a100= 601, a99= 595, a98= 589, a97 = 583, a96= 577, etc…

  8. So, now we have : (with thanks to Carl Friedrich Gauss) 7 + 13 + 19 + 25 + 31 + … + 577 + 583 + 589 + 595 + 601 Just for grins, let’s pair up the terms of the series. 7) What do you notice about each pair of numbers? SOLUTION : They all add up to 608. 8) How many pairs of numbers are there? Answer: 50

  9. 9) So, what do we have so far? SOLUTION : S100 = (50)(608) = 30,400 10) How did we get the 50 from the original information? Answer : 11) Where does the 608 first come from? SOLUTION : 7 + 601 or a1+ a100 So, a formula that works for this problem is

  10. So, an arithmetic seriesformula is Since an = a1 + (n – 1)d, we can substitute to get : So, a second arithmetic series formula is Note the green fill above. Ah, hem, aren’t you supposed to be doing something now?

  11. Find the sum of the arithmetic series. Answer : 1568 12) Answer : -775 13) 14) Find the value of n. Answer : i = 46

  12. For more information : http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut54d_geom.htm http://www.math.montana.edu/frankw/ccp/general/sigma/learn.htm http://quiz.econ.usyd.edu.au/mathquiz/sigma/index.php http://www.mathwords.com/a/arithmetic_series.htm

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