11.1 Arithmetic Sequences and Series
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1 11.1 Arithmetic Sequeces ad Series
2 A itroductio 1, 4, 7, 10, 13 9, 1, 7, 15 6., 6.6, 7, 7.4 ππ+, 3, π+ 6 Arithmetic Sequeces ADD To get ext term π + 9, 4, 8, 16, 3 9, 3, 1, 1/ 3 1,1/ 4,1/16,1/ 64 π,.5 π,6.5π Geometric Sequeces MULTIPLY To get ext term 6 0 / 3 85 / π Arithmetic Series Sum of Terms Geometric Series Sum of Terms
3 Fid the ext four terms of 9, -, 5, Arithmetic Sequece 9 = 5 = 7 7 is referred to as the commo differece (d) Commo Differece (d) what we ADD to get ext term Next four terms 1, 19, 6, 33
4 Fid the ext four terms of 0, 7, 14, Arithmetic Sequece, d = 7 1, 8, 35, 4 Fid the ext four terms of x, x, 3x, Arithmetic Sequece, d = x 4x, 5x, 6x, 7x Fid the ext four terms of 5k, -k, -7k, Arithmetic Sequece, d = -6k -13k, -19k, -5k, -3k
5 Vocabulary of Sequeces (Uiversal) a1 a S d First term th term umber of terms sum of terms commo differece 1 ( ) th term of arithmetic sequece a = a + 1 d sum of terms of arithmetic sequece S = a + a ( ) 1
6 Give a arithmetic sequece with a 15 = 38 ad d = 3, fid a 1. x NA -3 a1 a S d First term th term umber of terms sum of terms commo differece 1 ( ) ( 5 )( 3) a = a + 1 d 38 = x X = 80
7 Fid S63 of 19, 13, 7, a1 First term 353?? 63 a th term umber of terms x 6 S d sum of terms commo differece 1 ( ) a = a + 1 d ( 3 )( 6)?? = ?? = 353 S = ( a1+ a) 63 S63 = S63 = 1051 ( )
8 Try this oe: Fid a16 if a1 = 1.5 ad d = a1 First term x 16 NA 0.5 a S d th term umber of terms sum of terms commo differece 1 ( ) a = a + 1 d ( ) a16 = a16 = 9
9 Fid if a = 633, a1 = 9, ad d = x NA 4 a1 a S d 1 First term th term umber of terms sum of terms commo differece ( ) a = a + 1 d ( 1) 633 = 9 + x = 9 + 4x 4 X = 7
10 Fid d if a1 = 6 ad a9 = NA x a1 a S d 1 First term th term umber of terms sum of terms commo differece ( ) a = a + 1 d ( 1) 0 = x 6 = 8x 13 x = 14
11 Fid two arithmetic meas betwee 4 ad NA x -4,,, 5 a1 a S d 1 First term th term umber of terms sum of terms commo differece ( ) ( 1)( ) a = a + 1 d 5 = 4+ 4 x x = 3 The two arithmetic meas are 1 ad, sice 4, -1,, 5 forms a arithmetic sequece
12 Fid three arithmetic meas betwee 1 ad NA x 1,,,, 4 a1 a S d 1 First term th term umber of terms sum of terms commo differece ( ) ( 5 1)( ) a = a + 1 d 4 = 1+ x 3 x = 4 The three arithmetic meas are 7/4, 10/4, ad 13/4 sice 1, 7/4, 10/4, 13/4, 4 forms a arithmetic sequece
13 Fid for the series i which a1 = 5, d = 3, S = y x a1 a S d 1 First term th term umber of terms sum of terms commo differece ( ) ( x 1) a = a + 1 d y = 5+ 3 S = a + a x 440 = ( y) 5 + ( ) 1 x 440 = ( x 1) 3 x( 7+ 3x) 440 = 880 = x 7 + 3x ( ) ( ) 0 = 3x + 7x 880 Graph o positive widow X = 16
14 1. Geometric Sequeces ad Series
15 1, 4, 7, 10, 13 9, 1, 7, 15 6., 6.6, 7, 7.4 ππ+, 3, π+ 6 Arithmetic Sequeces ADD To get ext term π + 9, 4, 8, 16, 3 9, 3, 1, 1/ 3 1,1/ 4,1/16,1/ 64 π,.5 π,6.5π Geometric Sequeces MULTIPLY To get ext term 6 0 / 3 85 / π Arithmetic Series Sum of Terms Geometric Series Sum of Terms
16 Vocabulary of Sequeces (Uiversal) a1 a S r First term th term umber of terms sum of terms commo ratio th term of geometric sequece a = ar 1 sum of terms of geometric sequece S 1 = ( ) a1 r 1 r 1
17 Fid the ext three terms of, 3, 9/,,, 3 vs. 9/ 3 ot arithmetic 3 9/ 3 = = 1.5 geometric r = , 3,,,, , 3,,, 4 8, 16
18 If a 1 1 =, r =, fid a 9. 3 a1 a S r First term th term umber of terms sum of terms commo ratio 1/ x 9 NA /3 a = 1 ar x = x = 8 = =
19 Fid two geometric meas betwee ad 54 -,,, 54 a1 a S r First term th term umber of terms sum of terms commo ratio NA x a = 1 ar 1 ( )( ) 4 54 = x 7 = x 3 = x 3 1 The two geometric meas are 6 ad -18, sice, 6, -18, 54 forms a geometric sequece
20 Fid a a if a = 3 ad r = ,,, Sice r = ,,, a a4 = = 9 9
21 Fida9 of,,,... a1 a S r First term th term umber of terms sum of terms commo ratio a = 1 ar 1 x 9 NA r = = = x = ( ) 9 1 x = ( ) 8 x = 16
22 If a5 = 3 ad r =, fid a a1 a S r,,,,3 First term th term umber of terms sum of terms commo ratio a = 1 ar 1 x 3 5 NA ( ) = x 3 = x ( ) 4 3 = 4x 8 = x
23 *** Isert oe geometric mea betwee ¼ ad 4*** *** deotes trick questio a1 a 1,,4 4 First term th term 1/4 4 umber of terms 3 S r sum of terms commo ratio NA x 4 = 1 4 r = 1 4 r a = 1 ar 1 16 = r ± 4 = r 1,1,4 4 1, 1,4 4
24 1 1 1 Fid S7 of a1 First term 1/ a th term NA S = S ( ) a1 r 1 r 1 r umber of terms sum of terms commo ratio 7 x r = = = x = = = 63 64
25 Sectio 1.3 Ifiite Series
26 1, 4, 7, 10, 13,. Ifiite Arithmetic No Sum S = a + a 3, 7, 11,, 51 Fiite Arithmetic ( ) 1,, 4,, 64 Fiite Geometric S 1 = 1 ( ) a r 1 r 1 1,, 4, 8, Ifiite Geometric r > 1 r < -1 No Sum ,1,,, Ifiite Geometric -1 < r < 1 a S = 1 1 r
27 1 1 1 Fid the sum, if possible: r = = = 1 r 1Yes 1 1 = a 1 1 r = 1 = 1 1 S
28 Fid the sum, if possible: r = = 8 = 1 r 1 No NO SUM
29 Fid the sum, if possible: r = = = 1 r 1Yes S a r = = =
30 Fid the sum, if possible: r = 7 = 7 = 1 r 1No NO SUM
31 5 Fid the sum, if possible: r = = = 1 r 1Yes 10 5 = a 10 1 r = 1 = 1 1 S 0
32 The Boucig Ball Problem Versio A A ball is dropped from a height of 50 feet. It rebouds 4/5 of it s height, ad cotiues this patter util it stops. How far does the ball travel? / / S= + =
33 The Boucig Ball Problem Versio B A ball is throw 100 feet ito the air. It rebouds 3/4 of it s height, ad cotiues this patter util it stops. How far does the ball travel? / / S= + =
34 Sigma Notatio
35 UPPER BOUND (NUMBER) SIGMA (SUM OF TERMS) B = A a NTH TERM (SEQUENCE) LOWER BOUND (NUMBER)
36 4 ( j + ) = ( 1+ ) + ( + ) + ( 3+ ) ( 4 ) j= = 18 7 ( a) = ( ( 4) ) + ( 5 ( )) + ( 6 ( )) ( 7) a= 4 ( ) + = 44 4 ( ) ( 0 = ) ( ) ( ) ( ) ( ) = 0 = 33.5
37 b= = a S = 1 = 6 1 r 3 = ( x + 1) ( ) x= 7 ( 7 1) ( ( 8) 1) ( ( 9) 1 )... ( ( 3) 1) = S = ( a1+ a) = ( ) = 57
38 19 ( 4b + 3) ( ) b= 4 ( 4 4 3) ( 4( 5) 3) ( 4( 6) 3 )... ( 4( 19) 3) = S = ( a1+ a) = ( ) = 784
39 Rewrite usig sigma otatio: Arithmetic, d= 3 1 ( ) a = a + 1 d ( ) a = a = 3 4 = 1 3
40 Rewrite usig sigma otatio: Geometric, r = ½ 1 a = ar 1 1 a = 16 5 =
41 Rewrite usig sigma otatio: Not Arithmetic, Not Geometric a = 0 5 =
42 Rewrite the followig usig sigma otatio: Numerator is geometric, r = 3 Deomiator is arithmetic d= 5 NUMERATOR: ( ) a = 3 3 DENOMINATOR: ( ) a = a = 5 SIGMA NOTATION: = 1 ( )
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