suatu deret geometri dengan s=8, jumlah suku-suku genapnya =2. tentukanlah suku pertama dan rasionya

S = 8
a + ar + ar^2 + ... = 8
a/(1-r) = 8
a = 8-8r

Sgenap = 2
ar + ar^3 + ar^5 + ... = 2
ar/(1-r^2) = 2
(8-8r)r = 2(1-r^2)
8r-8r^2 = 2-2r^2
6r^2-8r+2=0
(r-1)(6r-2) = 0
r = 1 (Tidak mungkin)
r = 2/6 = 1/3
Maka,
a = 8-8r
a = 8 - 8/3 = 16/3
Sehingga,
a = 16/3 dan r = 1/3