tentukan jumlah deret geometri 4/9+4/3+4+...+108

Kelas 8 Matematika
Bab Barisan dan Deret Bilangan

4/9 + 4/3 + 4 + ... + 108

r = U2/U1
r = 4/3 : 4/9
r = 4/3 . 9/4
r = 3

Un = 108
a . r^(n - 1) = 108
4/9 . 3^(n - 1) = 108
3^(n - 1) = 108 . 9/4
3^(n - 1) = 243
3^(n - 1) = 3^5
n - 1 = 5
n = 5 + 1
n = 6

Sn = a . (r^n - 1) / (r - 1)
S6 = 4/9 . (3^6 - 1) / (3 - 1)
S6 = 4/9 . (729 - 1)/2
S6 = 4/9 . 728/2
S6 = 1.456/9
S6 = 161 7/9