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Barisan dan deret 1 bilingual

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Barisan dan deret 1 bilingual

  1. 1. KONSEP BARISAN DAN DERET
  2. 2. THE CONCEPT OF SEQUENCE AND SERIES
  3. 3. Hal.: 3 BARISAN DAN DERET AdaptifHal.: 3 Pola Barisan dan Deret Bilangan Kompetensi Dasar : Menerapkan konsep barisan dan deret aritmatika Indikator : 1. Nilai suku ke- n suatu barisan aritmatika ditentukan menggunakan rumus 2. Jumlah n suku suatu deret aritmatika ditentukan dengan menggunakan rumus
  4. 4. Hal.: 4 BARISAN DAN DERET AdaptifHal.: 4 The Pattern of Sequence and Series Number Basic Competence: Applying the concept of arithmetic sequence and series Indicator : 1. The value of n-th term in an arithmetic sequence is defined by formula 2. The sum of n in term of arithmetic sequence is defined by formula
  5. 5. Hal.: 5 BARISAN DAN DERET AdaptifHal.: 5 Saat mengendarai motor, pernahkah kalian mengamati speedometer pada motor tersebut? Pada speedometer terdapat angka-angka 0,20, 40, 60, 80, 100, dan 120 yang menunjukkan kecepatan motor saat kalian mengendarainya. Angka-angka ini berurutan mulai dari yang terkecil ke yang terbesar dengan pola tertentu sehingga membentuk sebuah pola barisan Pola Barisan dan Deret Bilangan
  6. 6. Hal.: 6 BARISAN DAN DERET AdaptifHal.: 6 When you ride a motor cycle, have you ever look at the speeedometer? In speedometer,there are numbers of 0,20, 40, 60, 80, 100, and 120 which show the speed of your motor cycle. These numbers are un order, starts from the smallest to the biggest with certain pattern, so that it forms a pattern of sequence The Pattern of Sequence and Series Number
  7. 7. Hal.: 7 BARISAN DAN DERET AdaptifHal.: 7 Bayangkan anda seorang penumpang taksi. Dia harus membayar biaya buka pintu Rp 15.000 dan argo Rp 2.500 /km. 15.000 17.500 20.000 22.500 ……. Buka pintu 1 km 2 km 3 km 4 km Pola Barisan dan Deret Bilangan
  8. 8. Hal.: 8 BARISAN DAN DERET Adaptif Imagine that you are a taxi passenger. You have to pay the starting fee Rp 15.000 and it charge Rp 2.500 /km. 15.000 17.500 20.000 22.500 ……. Starting fee 1 km 2 km 3 km 4 km The Pattern of Sequence and Series Number
  9. 9. Hal.: 9 BARISAN DAN DERET Adaptif NOTASI SIGMA Konsep Notasi Sigma Perhatikan jumlah 6 bilangan ganjil pertama berikut: 1 + 3 + 5 + 7 + 9 + 11 ……….. (1) Pada bentuk (1) Suku ke-1 = 1 = 2.1 – 1 Suku ke-2 = 3 = 2.2 – 1 Suku ke-3 = 5 = 2.3 – 1 Suku ke-4 = 7 = 2.4 – 1 Suku ke-5 = 9 = 2.5 – 1 Suku ke-6 = 11 = 2.6 – 1 Secara umum suku ke-k pada (1) dapat dinyatakan dalam bentuk 2k – 1, k ∈ { 1, 2, 3, 4, 5, 6 }
  10. 10. Hal.: 10 BARISAN DAN DERET Adaptif SIGMA NOTATION The Concept of Sigma Notation Look at the sum of the first sixth odd number below: 1 + 3 + 5 + 7 + 9 + 11 ……….. (1) In the form(1) The 1st term = 1 = 2.1 – 1 The 2nd term= 3 = 2.2 – 1 The 3rd term = 5 = 2.3 – 1 The 4th term = 7 = 2.4 – 1 The 5th term = 9 = 2.5 – 1 The 6th term = 11 = 2.6 – 1 Generally, the k-th term in (1) can be stated in the form of 2k – 1, k ∈ { 1, 2, 3, 4, 5, 6 }
  11. 11. Hal.: 11 BARISAN DAN DERET Adaptif NOTASI SIGMA Dengan notasi sigma bentuk penjumlahan (1) dapat ditulis : ∑= =+++++ 6 1k 1)-(2k1197531
  12. 12. Hal.: 12 BARISAN DAN DERET Adaptif SIGMA NOTATION In Sigma notation, the addition form (1) can be written as: ∑= =+++++ 6 1k 1)-(2k1197531
  13. 13. Hal.: 13 BARISAN DAN DERET Adaptif Bentuk ∑ = − 6 1 )12( k k dibaca “sigma 2k – 1 dari k =1 sampai dengan 6” atau “jumlah 2k – 1 untuk k = 1 sd k = 6” 1 disebut batas bawah dan 6 disebut batas atas, k dinamakan indeks (ada yang menyebut variabel) ∑ = −− 9 4 )1)3(2( k k ∑ = − 9 4 )72( k k NOTASI SIGMA
  14. 14. Hal.: 14 BARISAN DAN DERET Adaptif In the form of ∑ = − 6 1 )12( k k It is read “sigma 2k – 1 from k =1 to 6” or “the sum of 2k – 1 for k = 1 sd k = 6” 1 is called lower limit and 6 is called upper limit, k is called index (some people called it variable) ∑ = −− 9 4 )1)3(2( k k ∑ = − 9 4 )72( k k SIGMA NOTATION
  15. 15. Hal.: 15 BARISAN DAN DERET Adaptif NOTASI SIGMA Secara umum
  16. 16. Hal.: 16 BARISAN DAN DERET Adaptif SIGMA NOTATION Generally
  17. 17. Hal.: 17 BARISAN DAN DERET Adaptif Nyatakan dalam bentuk sigma 1. a + a2 b + a3 b2 + a4 b3 + … + a10 b9 ∑ − = 10 1k )1kbk(a )142()132()122()112()12( 4 1 +⋅++⋅++⋅++⋅=+∑=k k Contoh: 249753 =+++= Hitung nilai dari: NOTASI SIGMA
  18. 18. Hal.: 18 BARISAN DAN DERET Adaptif Stated into sigma form 1. a + a2 b + a3 b2 + a4 b3 + … + a10 b9 ∑ − = 10 1k )1kbk(a )142()132()122()112()12( 4 1 +⋅++⋅++⋅++⋅=+∑=k k Example: 249753 =+++= Define the value of SIGMA NOTATION
  19. 19. Hal.: 19 BARISAN DAN DERET Adaptif NOTASI SIGMA nn n 1n bCabC...baCbaCbaCa n 1n 33nn 3 22nn 2 1nn 1 n ++++++ − − −−− ∑ = − n 0r rrnn r baC 2. (a + b)n =
  20. 20. Hal.: 20 BARISAN DAN DERET Adaptif SIGMA NOTATION nn n 1n bCabC...baCbaCbaCa n 1n 33nn 3 22nn 2 1nn 1 n ++++++ − − −−− ∑ = − n 0r rrnn r baC 2. (a + b)n =
  21. 21. Hal.: 21 BARISAN DAN DERET Adaptif Sifat-sifat Notasi Sigma : , Untuk setiap bilangan bulat a, b dan n .....1 321 1 n n k aaaaak +++=∑= ∑∑ == = n mk n mk akCCak.2 ∑∑∑ === +=+ n mk n mk n mk bkakbkak )(.3 ∑∑ + +== − pn pmk n mk pakak.4 CmnC n mk )1(.5 +−=∑= ∑∑∑ == − = =+ n mk n pk p mk akakak 1 .6 0.7 1 =∑ = = m mk ak NOTASI SIGMA
  22. 22. Hal.: 22 BARISAN DAN DERET Adaptif The properties of sigma notation : , For every integer a, b and n .....1 321 1 n n k aaaaak +++=∑= ∑∑ == = n mk n mk akCCak.2 ∑∑∑ === +=+ n mk n mk n mk bkakbkak )(.3 ∑∑ + +== − pn pmk n mk pakak.4 CmnC n mk )1(.5 +−=∑= ∑∑∑ == − = =+ n mk n pk p mk akakak 1 .6 0.7 1 =∑ = = m mk ak SIGMA NOTATION
  23. 23. Hal.: 23 BARISAN DAN DERET Adaptif NOTASI SIGMA Contoh1: Tunjukkan bahwa Jawab : ∑∑ == +=+ 3 1 3 1 )24()24( jk ji 30)33.4()22.4()21.4()24( 3 1 =+++++=+∑=i i 30)23.4()22.4()21.4()24( 3 1 =+++++=+∑=j j
  24. 24. Hal.: 24 BARISAN DAN DERET Adaptif SIGMA NOTATION Example 1: Show that Answer : ∑∑ == +=+ 3 1 3 1 )24()24( jk ji 30)33.4()22.4()21.4()24( 3 1 =+++++=+∑=i i 30)23.4()22.4()21.4()24( 3 1 =+++++=+∑=j j
  25. 25. Hal.: 25 BARISAN DAN DERET Adaptif NOTASI SIGMA ∑∑ == + 6 4 2 3 1 2 66 kk kk ∑∑∑∑ ==== ==+ 6 1 2 6 1 2 6 4 2 3 1 2 6666 kkkk kkkk Hitung nilai dari Contoh 2 : Jawab: = 6 (12 +22 + 32 + 42 + 52 + 62 ) = 6 (1 + 4 + 9 + 16 + 25 + 36) = 6.91 = 546
  26. 26. Hal.: 26 BARISAN DAN DERET Adaptif SIGMA NOTATION ∑∑ == + 6 4 2 3 1 2 66 kk kk ∑∑∑∑ ==== ==+ 6 1 2 6 1 2 6 4 2 3 1 2 6666 kkkk kkkk Define the value of Example 2 : Answer: = 6 (12 +22 + 32 + 42 + 52 + 62 ) = 6 (1 + 4 + 9 + 16 + 25 + 36) = 6.91 = 546
  27. 27. Hal.: 27 BARISAN DAN DERET Adaptif BARISAN DAN DERET ARITMATIKA  Bilangan-bilangan berurutan seperti pada speedometer memiliki selisih yang sama untuk setiap dua suku berurutannya sehingga membentuk suatu barisan bilangan  Barisan Aritmatika adalah suatu barisan dengan selisih (beda) dua suku yang berurutan selalu tetap Bentuk Umum : U1, U2, U3, …., Un a, a + b, a + 2b,…., a + (n-1)b Pada barisan aritmatika,berlaku Un – Un-1 = b sehingga Un= Un-1 + b
  28. 28. Hal.: 28 BARISAN DAN DERET Adaptif ARITHMETIC SEQUENCE AND SERIES  The orderly numbers like in speedometer have the same difference for every two orderly term, so it forms a sequence  Arithmetic sequence is sequence with difference two orderly term constant  The general form is : U1, U2, U3, …., Un a, a + b, a + 2b,…., a + (n-1)b  In arithmetic sequence, we have Un – Un-1 = b, so Un= Un-1 + b
  29. 29. Hal.: 29 BARISAN DAN DERET Adaptif BARISAN DAN DERET ARITMATIKA
  30. 30. Hal.: 30 BARISAN DAN DERET Adaptif If you start arithmetic sequence with the first term a and difference b, then you will get this following sequence The n-th term of arithmetic sequence is Un = a + ( n – 1 )b Where Un = n-th term a = the first term b = difference n = the term’s quantity ARITHMETIC SEQUENCE AND SERIES a a + b a + 2b a + 3b …. a + (n-1)b
  31. 31. Hal.: 31 BARISAN DAN DERET AdaptifHl.: 31 BARISAN DAN DERET ARITMATIKA
  32. 32. Hal.: 32 BARISAN DAN DERET Adaptif If every term of arithmetic sequence is added, then we will get arithmetic series. Arithmetic series is the sum of terms of arithmetic sequence General form : U1 + U2 + U3 + … + Un atau a + (a +b) + (a+2b) +… + (a+(n-1)b) The formula of the sum of the first term in arithmetic series is Where S = the sum of n-th term n = the quantity of term a = the first term b = difference = n-th term ARITHMETIC SEQUENCE AND SERIES ( )bna n Sn )1(2 2 −+=
  33. 33. Hal.: 33 BARISAN DAN DERET Adaptif BARISAN DAN DERET ARITMATIKA
  34. 34. Hal.: 34 BARISAN DAN DERET Adaptif Known: the sequence of 5, -2, -9, -16,…., find: a.The formula of n-th term b.The 25th term Answer: The difference of two orderly terms in sequence 5,-2, -9,-16 ,…is constant, b= -7, so that the sequence is an arithmetic sequence a.The formula of the n-th term in arithmetic sequence is Un = 5 + ( n – 1 ). -7 Un = 5 + - 7n + 7 Un = -7n + 12 b. The 25th term of arithmetic sequence is : U12 = - 7.12 + 12 = - 163 ARITHMETIC SEQUENCE AND SERIES
  35. 35. Hal.: 35 BARISAN DAN DERET Adaptif Barisan geometri adalah suatu barisan dengan pembanding (rasio) antara dua suku yang berurutan selalu tetap. Ada selembar kertas biru, akan dipotong-potong menjadi dua bagian. BARISAN DAN DERET GEOMETRI
  36. 36. Hal.: 36 BARISAN DAN DERET Adaptif Geometric sequence is a sequence which has the constant ratio between two orderly term There is blue paper. It will cut into two pieces GEOMETRIC SEQUENCE AND SERIES
  37. 37. Hal.: 37 BARISAN DAN DERET Adaptif BARISAN DAN DERET GEOMETRI
  38. 38. Hal.: 38 BARISAN DAN DERET Adaptif Look at the paper part that form a sequence Every two orderly terms of the sequence have the same ratio It seems that the ratio of every two orderly terms in the sequence is always constant. The sequence like this is called geometric sequence and the comparison of every two orderly term is called ratio (r) 1 2 4 U1 U2 U3 2.... 12 3 1 2 ==== −n n U U U U U U GEOMETRIC SEQUENCE AND SERIES
  39. 39. Hal.: 39 BARISAN DAN DERET AdaptifHal.: 39 BARISAN DAN DERET GEOMETRI
  40. 40. Hal.: 40 BARISAN DAN DERET Adaptif Geometric sequence is a sequence which have constant ratio for two orderly term General form: U1, U2, U3, …., Un atau a, ar, ar2 , …., arn-1 In geometric sequence If you start the geometric sequence with the first term a and the ratio is r, then you get the following sequence GEOMETRIC SEQUENCE AND SERIES r U U n n = −1 1. −= nn UrsehinggaU
  41. 41. Hal.: 41 BARISAN DAN DERET Adaptif Suku ke-n barisan Geometri adalah : BARISAN DAN DERET GEOMETRI
  42. 42. Hal.: 42 BARISAN DAN DERET Adaptif The n-th term of geometric sequence is : GEOMETRIC SEQUENCE AND SERIES Start With the first term a Multiply with ratio r Write the multiplication result
  43. 43. Hal.: 43 BARISAN DAN DERET Adaptif BARISAN DAN DERET GEOMETRI Hubungan suku-suku barisan geometri Seperti dalam barisan Aritmatika hubungan antara suku yang satu dan suku yang lain dalam barisan geometri dapat dijelaskan sebagai berikut: Ambil U12 sebagai contoh : U12 = a.r11 U12 = a.r9 .r2 = U10. r2 U12 = a.r8 .r3 = U9. r3 U12 = a.r4 .r7 = U5. r7 U12 = a.r3 .r8 = U4.r8 Secara umum dapat dirumuskan bahwa : Un = Uk. rn-k
  44. 44. Hal.: 44 BARISAN DAN DERET Adaptif GEOMETRIC SEQUENCE AND SERIES The relation of terms in geometric sequence Like in arithmetic sequence, the relation between terms in geometric sequence can be explained as follows: Take U12 as example : U12 = a.r11 U12 = a.r9 .r2 = U10. r2 U12 = a.r8 .r3 = U9. r3 U12 = a.r4 .r7 = U5. r7 U12 = a.r3 .r8 = U4.r8 Generally, it can be formulated Un = Uk. rn-k
  45. 45. Hal.: 45 BARISAN DAN DERET Adaptif BARISAN DAN DERET GEOMETRI
  46. 46. Hal.: 46 BARISAN DAN DERET Adaptif GEOMETRIC SEQUENCE AND SERIES Geometric series is the sum of terms in geometric sequence General form U1 + U2 + U3 + …. + Un a + ar + ar2 + ….+ arn-1 The formula of the n sum of the first term in geometric series is 1, 1 )1( < − − = r r ra S n n
  47. 47. Hal.: 47 BARISAN DAN DERET Adaptif BARISAN DAN DERET GEOMETRI
  48. 48. Hal.: 48 BARISAN DAN DERET Adaptif GEOMETRIC SEQUENCE AND SERIES Known sequence 27, 9, 3, 1, …..find a.The formula of the n-th term b. The 8th term Answer: The ratio of two orderly terms in sequence 27,9,3, 1, …is constant, so that the sequence is a geometric sequence a. The formula of the n-th term in geometric sequence is 3 1 =r 1 3 1 27 −       = n nU 113 )3.(3 −− = n 13 3.3 +− = n n− = 4 3
  49. 49. Hal.: 49 BARISAN DAN DERET Adaptif GEOMETRIC SEQUENCE AND SERIES b. The 8th term of geometric sequence is 84 8 3 − =U 4 3− = 81 1 = n nU − = 4 3
  50. 50. Hal.: 50 BARISAN DAN DERET Adaptif Deret geometi tak hingga adalah deret geometri yang banyak suku- sukunya tak hingga. Jika deret geometri tak hingga dengan -1 < r < 1 , maka jumlah deret geometri tak hingga tersebut mempunyai limit jumlah (konvergen). Untuk n = ∞ , rn mendekati 0 Sehingga S∞ = Dengan S∞ = Jumlah deret geometri tak hingga a = Suku pertama r = rasio Jika r < -1 atau r > 1 , maka deret geometri tak hingganya akan divergen, yaitu jumlah suku-sukunya tidak terbatas Deret Geometri tak hingga r a −1 r ra Sn n − − = 1 )1( BARISAN DAN DERET GEOMETRI
  51. 51. Hal.: 51 BARISAN DAN DERET Adaptif Infinite geometric series is a geometric series which has infinite terms. If infinite geometric series is -1 < r < 1 , then the sum of geometric series has sum limit (convergent). For n = ∞ , rn is close to 0 So S∞ = With S∞ = the sum of infinite geometric series a = the first term r = ratio If r < -1 or r > 1 , then the infinite geometric series will be divergent, means the sum of terms is not limited Infinite Geometric Series r a −1 r ra Sn n − − = 1 )1( GEOMETRIC SEQUENCE AND SERIES
  52. 52. Hal.: 52 BARISAN DAN DERET Adaptif 1. Hitung jumlah deret geometri tak hingga : 18 + 6 + 2 + … . . Contoh : 3 1 2 3 1 2 === u u u u r BARISAN DAN DERET GEOMETRI 27 3 2 18 3 1 1 18 1 == − = − =∞ r a s Jawab : a = 18 ;
  53. 53. Hal.: 53 BARISAN DAN DERET Adaptif 1. Find the sum of infinite geometric series : 18 + 6 + 2 + … . . Example : 3 1 2 3 1 2 === u u u u r GEOMETRIC SEQUENCE AND SERIES 27 3 2 18 3 1 1 18 1 == − = − =∞ r a s Answer : a = 18 ;
  54. 54. Hal.: 54 BARISAN DAN DERET Adaptif 2. Sebuah bola elastis dijatuhkan dari ketinggian 2m. Setiap kali memantul dari lantai, bola mencapai ketinggian ¾ dari ketinggian sebelumnya. Berapakah panjang lintasan yang dilalui bola hingga berhenti ? BARISAN DAN DERET GEOMETRI Lihat gambar di samping! Bola dijatuhkan dari A, maka AB dilalui satu kali, selanjutnya CD, EF dan seterusnya dilalui dua kali. Lintasannya membentuk deret geometri dengan a = 3 dan r = ¾ Panjang lintasan = 2 S∞ - a 2 4 1 2 2 2 4 3 1 2 2 1 2 −             = −             − = −      − = a r a = 14 Jadi panjang lintasan yang dilalui bola adalah14 m
  55. 55. Hal.: 55 BARISAN DAN DERET Adaptif 2. An elastic ball is drop from the height of 2m. Every time it bounce from the floor, it has ¾ of the previous height. How long is the route that will be passed by the ball until it stop? GEOMETRIC SEQUENCE AND SERIES Look at the picture! The ball is drop from A, so AB is passed only once. Then CD, EF, etc is passed twice. The route is in geometric series with a = 3 and r = ¾ the length of the route is= 2 S∞ - a 2 4 1 2 2 2 4 3 1 2 2 1 2 −             = −             − = −      − = a r a = 14 So, the route length that pass by the ball is 14 m
  56. 56. Hal.: 56 BARISAN DAN DERET Adaptif
  57. 57. Hal.: 57 BARISAN DAN DERET Adaptif

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