diberikan suatu barisan geometri dengan U1 + U4 = 4 dan U2 + U5 = 12. Tentukan : a. rasionya dan suku pertama b. suku ke-3 c. rumus suku ke - n

[tex]U_1+U_4=4 \\ a+ar^3=4 \\ a(1+r^3)=4 \\ \\ U_2+U_5=12 \\ ar+ar^4=12 \\ ar(1+r^3)=12 \\ r\times a(1+r^3)=12 \\ r\times 4=12 \\ r=3 \\ \\ a(1+r^3)=4 \\ a(1+27)=4 \\ 28a = 4 \\ a=\frac{1}{7} \\ \\ U_3=ar^2=\frac{1}{7}\times 3^2=\frac{1}{7}\times 9=\frac{9}{7} \\ \\ \displaystyle U_n=ar^{n-1} \\ U_n=\frac{1}{7}\times3^{n-1} \\ U_n = \frac{1}{7}\times\frac{1}{3}\times 3^n \\ U_n=\frac{3^n}{21}[/tex]

U1 + U4 = 4—a +ar^3 = 4—a( 1+ r^3) = 4 U2 + U5 = 12—ar + ar^4 = 12—ar( 1 + r^3) = 12 a ( 1 + r^3)/ar ( 1 + r^3) = 4/12 a/ar = 4/12 1/r = 1/3 r = 3 ar,= 12 a × 3 = 12 a = 4 Jadi suku pertama = a = U1 = 4 Rasio = r = 3 Suku ke -3 = ar^2 = 4 × 3^2 = 4 × 9 = 36 Rumus suku ke - n = Un = 4 × (3)^n-1