tentukan 3 suku pertama suatu barisan geometri u4=135 dan r=3

u(4) = a x 3^3 = 135
a = 135/27
a = 5

u(1) = 5 x 3^0 = 0
u(2) = 5 x 3^1 = 15
u(3) = 5 x 3^2 = 45

u4 = ar^4-1
135 = a × (3)^3
a = 135/27 = 5
u1 = a = 5
U2= ar = 5 × 3 = 15
U3 = ar^2 = 5 × 3^2 = 45