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9-2 Arithmetic Sequences & Series.  When another famous mathematician was in first grade, his teacher asked the class to add up the numbers one through.

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Presentation on theme: "9-2 Arithmetic Sequences & Series.  When another famous mathematician was in first grade, his teacher asked the class to add up the numbers one through."— Presentation transcript:

1 9-2 Arithmetic Sequences & Series

2  When another famous mathematician was in first grade, his teacher asked the class to add up the numbers one through a hundred (1+2+3 etc., all the way up to 100).  Write out the teacher’s request in summation notation, then find the answer (no calculators!) Try to figure out an efficient way!

3  1 to 100

4  Find the sum from 3 to 1,000 or

5 TTTThe difference between consecutive terms is constant (or the same). TTTThe constant difference is also known as the common difference (d). (It’s also that number that you are adding everytime!)

6  -10,-6,-2,0,2,6,10,…  -6--10=4  -2--6=4  0--2=2  2-0=2  6-2=4  10-6=4 Not arithmetic (because the differences are not the same)  5,11,17,23,29,…  11-5=6  17-11=6  23-17=6  29-23=6  Arithmetic (common difference is 6)

7 5, 8, 11, 14, 17, 20, 23…

8 a n =a 1 +(n-1)d a 1 2 variables need to be known (or solved for): a 1 and d Kind of like in y = mx+b, we need to know m and b a n = d(n-1)+a 1 a n = d(n-1)+a 1

9  The is a common difference where d=15, therefore the sequence is arithmetic.  Use a n =a 1 +(n-1)d a n =32+(n-1)(15) a n =32+(n-1)(15) a n =32+15n-15 a n =32+15n-15 a n =17+15n a n =17+15n a 12 =17+15(12)=197

10  Use a n =a 1 +(n-1)d to find the 1 st term! a 8 =a 1 +(8-1)(.25) 50=a 1 +(7)(.25) 50=a 1 +1.75 48.25=a 1 * Now, use a n =a 1 +(n-1)d to find the rule. a n =48.25+(n-1)(.25) a n =48.25+.25n-.25 a n =48+.25n This is like being given a slope and a (x,y) coordinate. We need to find the “b”!

11  Begin by writing 2 equations; one for each term given. a 5 =a 1 +(5-1)d OR 10=a 1 +4d And a 30 =a 1 +(30-1)d OR 110=a 1 +29d  Now use the 2 equations to solve for a 1 & d. 10=a 1 +4d 10=a 1 +4d 110=a 1 +29d (subtract the equations to cancel a 1 ) -100= -25d So, d=4 and a 1 =-6 (now find the rule) a n =a 1 +(n-1)d a n =-6+(n-1)(4) OR a n =-10+4n This is like being given 2 coordinates. We have to find the “slope” and the “b”

12 -2=-10+4n8=4n2=n

13  2+4+6+8+10+12  2+4+6+8+10+12+14+16  2+4+6+8+10+12+14+16…  Think of the story of Gauss adding 1 to 100 (12+2)(6/2) = 42 (16+2)(8/2) = 72

14  The sum of the terms in an arithmetic sequence  The formula to find the sum of a finite arithmetic series is: # of terms 1 st Term Last Term

15  Find the sum of the 1 st 25 terms.  We know the 1 st term, we need the 25 th term.  a n =20+(n-1)(-2)  a n =22-2n  So, a 25 = -28 (last term) Find n such that S n =-760

16 -1520=n(20+22-2n) -1520=-2n 2 +42n 2n 2 -42n-1520=0 n 2 -21n-760=0 (n-40)(n+19)=0 n=40 or n=-19 Always choose the positive solution!

17  Your book refers to partial sums of an arithmetic sequence. To find the nth partial sum… simply find the sum of the first n terms.  Example: To find the 50 th partial sum, find the sum of the first 50 terms.

18  Consider a job offer with a starting salary of $32,500 and an annual raise of $2500. Determine the total compensation from the company through the first ten years of employment.

19

20 9-2 Pg. 659 #3-47 odd, 57, 58, 81, 82


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