Tentukan jumlah 5 suku yang pertama dari deret geometri dari 81,27,9

a = 81
r = 1/3

Sn = a(1-r^n)/1-r
S5 = 81(1-(1/3^5)) / 1-1/3
S5 = 81 (1-1/243) / 2/3
S5 = 81 (242/243) / 2/3
S5 = 121

Semoga membantu

81, 27, 9,....
suku pertama a= 81
rasio r = 27/81 = 9/27 = 1/3
Sn = a (1 - r^n)/( 1- r)
S5 = 81 (1 - 1/3^5) / (1 - 1/3)
= 81 (1 - 1/243) / (2/3)
= 81 ( 242/243) / (2/3)
= 242/3 /(3/2)
= 242/2
= 121