Kuis [+50]:

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Deret geometri. Jika suku ke-6 adalah 40/81, dan suku ke-3 adalah 13⅓.
Berapa jumlah 8 suku pertama?

Kuis [+50]:

Jawaban:

U3 = 13⅓ = 40/3

U6 = 40/81

r (ratio)

U6 ÷ U3 = 40/81 ÷ 40/3

(a × r^(n – 1) ÷ (a × r^(n – 1)) = 40/81 × 3/40

r^(6 – 1) ÷ r^(3 – 1) = 3/81

r⁵ ÷ r² = 1/27

r³ = 1/27

r = 1/3

a (suku awal)

Un = a × r^(n – 1)

40/3 = a × r²

40/3 = a × (⅓)²

40/3 = a × 1/9

a = 40/3 ÷ 1/9

a = 40/3 × 9

a = 120

S8 (Jumlah 8 suku pertama)

Sn = (a × (1 – rⁿ)) / (1 – r)

S8 = (120 × (1 – (⅓)⁸) / (1 – ⅓)

S8 = (120 × (1 – (1/6561)) / (3/3 – ⅓)

S8 = (120 × (6561/ 6561 – 1/6561)) / ⅔

S8 = 120 × 6560/6561 / ⅔

S8 = 787.200/6561 / ⅔

S8 = 787.200/6561 × 3/2

S8 = 393.600/2.187

S8 = 131.200/729

S8 = 179 ⁷⁰⁹/729