Diketahui suku barisan geometri suku ke-1 = 2/3 dan suku ke- 3 = 2/27. Jumlah empat suku pertama S4 adalah ...

Jawaban : 80/81 Ingat! r = Un : (Un-1) dengan r = rasio Un = suku ke-n U(n-1) = suku ke - (n-1) Mencari suku ke -n (Un) Un = arⁿ⁻¹ Mencari jumlah n suku pertama (Sn) Sn = a (rⁿ - 1)/(r-1) jika r > 1 Sn = a (1-rⁿ) / (1-r) jika r < 1 Pembahasan U1 = a = 2/3 U3 = 2/27 U3 = 2/3 × r^(3-1) 2/27 = 2/3 × r² r² = 2/27 × 3/2 r² = 6/54 r² = (6 : 6)/(54 : 6)--> sederhanakan dibagi 6 r² = 1/9 r = ± √(1/9) r = ± √(1/3)² r = ± 1/3 Jumlah empat suku pertama S4 jika r = 1/3 Sn = a (1-rⁿ) / (1-r) jika r < 1 S4 = 2/3 (1 - (1/3)⁴)/(1 - 1/3) = 2/3 (1 - 1⁴/3⁴)/(3/3 - 1/3) = 2/3 (1 - 1/81)/(3-1)/3 = 2/3 (81/81 - 1/81) : 2/3 = (81 - 1)/81 = 80/81 jika r = 1/3 Sn = a (1-rⁿ) / (1-r) jika r < 1 S4 = 2/3 (1 - (1/3)⁴)/(1 - 1/3) = 2/3 (1 - 1⁴/3⁴)/(3/3 - 1/3) = 2/3 (1 - 1/81)/(3-1)/3 = 2/3 (81/81 - 1/81) : 2/3 = (81 - 1)/81 = 80/81 jika r= -1/3 S4 = 2/3 (1 - (-1/3)⁴)/(1 - (-1/3)) = 2/3 (1 - 1/81)/(1+1)/3 = 2/3 (81/81 - 1/81) : 2/3 = 2/3 × 3/2 × (81 - 1)/81 = 1 × 80/81 = 80/81 Jadi, Jumlah empat suku pertama S4 adalah 80/81