suatu deret geometri positif mempunyai suku ke2 bernilai 6 dan suku ke4 bernilai 54. jumlah delapan suku deret tersebut adalah...

U2 = ar = 6
U4 = ar^3 = 54

ar/ar^3 = 6/54
1/r^2 = 1/9
r = 3

ar = 6
a × 3 = 6
a = 2

S8 = a(r^n-1)/r-1
= 2(3^8-1)/2
= 3^8-1
= 6561-1
= 6560

U2 = 6
U4 = 54
n = 8

U2 = ar
6 = ar

U4 = ar^3
54 = ar^3

ar^3/ar = 54/6
r^2 = 9
r = V9
r = 3

ar = 6
a(3) = 6
a = 6/3
a = 2

Sn = a(1 - r^n)/(1 - r)
S8 = 2(1 - 3^8)/(1 - 3)
S8 = 2(1 - 6561)/(-2)
S8 = 2(-6560)/(-2)
S8 = -13120/(-2)
S8 = 6560

Jadi jumlah delapan suku deret tersebut adalah 6560