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DERET ARITMATIKA

DERET ARITMATIKA. Standar Kompetensi. Menerapkan konsep barisan dan deret dalam pemecahan masalah. Kompetensi Dasar. 2. Menerapkan konsep barisan dan deret aritmatika. Tujuan Pembelajaran. DERET ARITMATIKA.

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DERET ARITMATIKA

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  1. DERET ARITMATIKA

  2. Standar Kompetensi Menerapkan konsep barisan dan deret dalam pemecahan masalah

  3. Kompetensi Dasar 2. Menerapkan konsep barisan dan deret aritmatika

  4. Tujuan Pembelajaran

  5. DERET ARITMATIKA Deret aritmatika adalahjumlahbarisanaritmatikasampaisukuke-n (u1+u2+u3+…..+un) Deretaritmatikadisimbulkandengan “S” Sehingga Jumlahn suku pertama = Sn

  6. RumusJumlahn suku pertama deretaritmatika: Ket: Sn= jumlahke-n Un= sukuke-n a = sukuke-1 n = nomorsuku b = beda(selisihduasuku yang berurutan)

  7. Contoh 1: Hitungjumlah11sukupertamadarideretaritmatika3+ 7 + 11 + 15……. Penyelesaian: Diket: a = 3 b = 7– 3 = 4 n = 11 Ditya: S11= ….. ?

  8. Contoh 2: Tentukan jumlah semua bilangan asli antara 1 dan 100 yang habis dibagi 3! Penyelesaian: Bilangan asli antara 1 dan 100 yang habis di bagi 3: 3 + 6 + 9 + 12 + ... + 99 a = 3 b = 3 Un = 99 Un = a + (n – 1) b 99 = 3 + (n – 1) 3 99 = 3 + 3n -3 99 = 3n n = 99 3 = 33 Maka Sn = n (a + Un) 2 = 33 (3 + 99) 2 = 33 . 102 2 = 1683

  9. Soal: Tentukan banyaknya bilangan yang habis di bagi 5 antara 1 sampai dengan 100! Hitunglah jumlah 30 suku pertama dari deret 4 + 7 + 10 + 13 + ... ! Tentukan suku pertama dan beda dari deret aritmatika jika diketahui S15 = 150 dan U15 = 24.

  10. Penyelesaian: Diket: Bilangan yang habis di bagi 5 antara 1 – 100: 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 + 65 + 70 + 75 + 80 + 85 + 90 + 95 + 100 a = 5 b = 5 n = 20 Ditya: S20 = ... ? Jawab: Sn = n/2 (a + Un) S20 = 20/2 (5 + 100) = 10 . 105 = 1050

  11. Diket: 4 + 7 + 10 + 13 + ... a = 4 b = 3 n = 30 Ditya: S30 = ... ? Jawab: Sn = n/2 (2a + (n - 1) b) S30 = 30/2 (2 . 4 + (30 – 1) 3) = 15 (8 + 29 . 3) = 15 (8 + 87) = 15 . 95 = 1425

  12. Diket: S15 = 150 U15 = 24 Ditya: a = ... ? b = ... ? Jawab: Un = a + (n – 1) b U15 = a + (15 – 1) b 24 = a + 14 b .....(1) Sn = n/2 (2 a + (n – 1) b) S15 = 15/2 (2 a + (15 – 1) b) 150 = 15/2 (2 a + 14 b) 150 = 15 a + 105 b ....(2) a + 14 b = 24 ....(1) a + 14 . 2 = 24 a + 28 = 24 a = 24 – 28 a = - 4 Jadi a = - 4 dan b = 2 (1)... a + 14 b = 24 |X15| 15 a + 210 b = 360 (2)... 15 a + 105 b = 150|X 1 | 15 a + 105 b = 150 105 b = 210 b = 2 -

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