suatu deret geometri tak hingga jumlahnya 5/8 dan rasionya 1/5. tentukan suku pertama dan jumlah 4 suku pertama deret tersebut

S~ = a / 1 - r
5/8 = a / 1 - 1/5
5/8 = a / 4/5
5/8 = 5a/4
20 = 40a
a = 20/40 = 1/2 S4 = 1/2 [1 - (1/5)^4] / 1 - 1/5 = 1/2 (1 - 1/625) / 4/5 = 5/8 × 624/625 = 1/1 × 33/125 = 33/125

Bab Deret Tak Hingga
Matematika SMA Kelas XI

Geometri tak hingga
1] a = S~ x (1 - r)
a = 5/8 x (1 - 1/5)
a = 5/8 x 4/5
a = 1/2

2] S4 = a x (1 - r^n) / (1 - r)
S4 = 1/2 x (1 - (1/5)^4) / (1 - 1/5)
S4 = 1/2 x (625 - 1)/625 : 4/5
S4 = 1/2 x 5/4 x 624/625
S4 = 78/125