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