Suku kedua dan kelima suatu deret geometri berturut-turut 10 dan 10/27. Jumlah n suku pertama deret tersebut adalah

misal xn = suku ke-n,
r = rasio, dan a = suku awal
x2 = ar, x5 = a(r^4)
x2/x5 =
ar/a(r^4) = 10/(10/27) =
1/(r^3) = 27
1/27 = (r^3)
1/3 = r
a = ar/r
a = x2/(1/3) = 10/(1/3) = 30
rumus jumlah deret gometri dngn r < 1 =
(a(1-(r^n)))/(1-r)
= (30(1-(1/3)^n))/(1-(1/3))
= (30(1-(1/3)^n))/(2/3)
= 45(1-(1/3)^n)