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SEQUENCES

1. 5. 4. 3. 2. 10. 7. 1. SEQUENCES. 9. 2. 3. 4. 6. 5. 8. BY: CAHYO HENY MEILIANA K1310019 NABILLAH K1310000. DEFINITION Sequence is arrangement of number which has certain rules. Examples : 2 , 4, 6, 8, 10, 12, 14, …. 2 , 4, 8, 16, 32, 64, 128 ,….

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SEQUENCES

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  1. 1 5 4 3 2

  2. 10 7 1 SEQUENCES 9 2 3 4 6 5 8 • BY: • CAHYO HENY MEILIANA K1310019 • NABILLAH K1310000

  3. DEFINITION • Sequence is arrangement of number which has certain rules. • Examples : • 2, 4, 6, 8, 10, 12, 14, …. • 2, 4, 8, 16, 32, 64, 128,…. • 9, 3, 1/3, 1/9, 1/27, ….

  4. 1. Arithmetic progressions • Arithmetic progressions is sequence of numbers which has difference between two of ordered terms and always constant.Examples: • 1, 3, 5, 7, 9, … . • the difference is 2 • 10, 8, 6, 4, 2,… . • the difference is -2

  5. Generally, we can write : • U1’ U2, U3, U4,…. Un • If U1= a and difference between ordered terms = b, so that • a, a + b, a + 2b, a + 3b, a + 4b, …. . a + ( n – 1 ) b

  6. If first terms called a and difference of ordered term is called b, so: • U 1 ,U2 , U3 , U4, …, Un-2 , Un-1 , Un a, a+b, a+2b, a+3b,… , a+(n-3)b, a+(n-2)b, a+(n-1)b

  7. Term of - n • Un = a + ( n – 1 ) b Un = Term of – n a = first term b = difference of ordered term n = total of term

  8. Example: • Given Arithmetic progression 3, 9, 15, 21, ……Find term of 20 Solution: U20 = a + ( n – 1 ) b = 3 + ( 20 – 1 ) 6 = 3 + 19. 6 = 3 + 114 = 117

  9. The center term • U1, U2, U3 the center term is U2 U2 = a + b = ½ ( 2a + 2b ) = ½ ( a + a + 2b ) = ½ ( U1 + U3 ) 2. U1, U2, U3, U4, U5 • the center term is U3 U2 = a + 2b = ½ ( 2a + 4b ) = ½ ( a + a + 4b ) = ½ ( U1 + U5)

  10. etc… • U1, U2, U3, U4, U5 …. . Un. • Uc= ½ ( U1 + Un) • with n = odd

  11. Interpolated • If given 2 of numbers x and y will be interpolated a number of k so that formed arithmetic progression: x, x + bi, x + 2bi, x + 3bi, ….. x + kbi, y the difference of interpolated is bi= y – ( x + kbi) bi+ kbi= y – x bi = ( y – x )/(1 + k) bi = difference of interpolated y = second number x = first number k = total of interpolated

  12. Example: • Between 7 and 107 be interpolated 19 of numbers so that formed arithmetic progression. • Find: 1. difference of sequence that formed 2. Term of 10 3. The center term if exist

  13. Solution: 1. Difference of sequence Given x = 7 and y = 107 bs = ( y – x )/(k + 1) = (107 – 7 )/(19+1) = ( 100 ) / 20 = 100/20 = 5 2. Term of 10 U10= a + ( n – 1 ) bs = 7 + ( 10 – 1 ) 5 = 7 + 9.5 = 7 + 45 = 52 3. Un = 107 107 = 7 + ( n – 1 ) 5 100 = 5n – 5 105 = 5n n = 105 / 5 n = 21 Hence, Uc= ½ ( U1 + U21 ) = ½ ( 7 + 107 ) = ½ ( 114 ) = 57 Thus, the center term is 57

  14. Geometric Progression • Definition • Geometric Progression is sequence of number which has constant ratio . • U1, U2, U3, U4, U5, … Un

  15. U1, U2, U3, U4, U5, … Un • IfU1 = a  U1= a • U2= ar • U3= ar2 • U4= ar3 • …. • Un= arn-1 Example: Given a sequence 2, 4, 8, … Find U7 ! Solution : U7= ar7-1 U7 = 2.26 U7 = 2. 64 U7 = 128

  16. The center term • The center term of geometric progression can be wrote such as: • U1, U2, U3,so that he center term is U2 • U2 =

  17. Similarly, • U1, U2, U3, U4, … Uk • if k is odd so that the center term is • Uc= • where k = odd • Uc= the center term • Uk= the term of k ( end )

  18. Example: From term of geometric progression given U2=4 and U10=1024, then find five of sequence first term! Solution: U2 =a r2-1= a r =4 …………………………. 1) U10= a r 10-1 = a r 9 =1024 ……………... 2) From equation 1) , we obtain: a = 4/r ……………3)

  19. Substitute equation 3) to equation 2) and we have, • (4/r).r9 =1024 • 4r8 =1024 • r 8 =256 • r =2 • then substitute value of r =2 to equation 3) to get value of a = 2. • Thus, five of geometric progression first term is 2, 4, 8, 16, 32.

  20. THANKS FOR YOUR ATTENTION

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