Question

Jumlah suatu deret geometri tak hingga adalah 64. Jika suku ke-2 adalah 12, tentukan jumlah 5 suku pertamanya.

Answer 1

U2 = 12
a.r = 12
a = 12/r

S∞ = 64
a/(1-r) = 64
a = 64.(1-r)
12/r = 64-64r
12 = 64r-64r^2
64r^2-64r+12 = 0 (bagi 4 kedua ruas)
16r^2-16r+3 = 0
16r^2-4r-12r+3 = 0
4r(4r-1)-3(4r-1) = 0
(4r-3)(4r-1) = 0
r = 3/4 atau r = 1/4

Jika r = 3/4:
a = 12/(3/4)
a = 12.4/3
a = 16

S5 = a.(1-r^n)/(1-r)
S5 = 16.(1-(3/4)^5)/(1-3/4)
S5 = 16.(1-243/1024)/(1/4)
S5 = 16.(781/1024).4/1
S5 = 781/16 = 48 13/16

S5 = a.(1-r^n)/(1-r)
S5 = 16.(1-(3/4)^5)/(1-3/4)
S5 = 16.(1-243/1024)/(1/4)
S5 = 16.(781/1024).4/1
S5 = 781/16 = 48 13/16

Jika r = 1/4:
a = 12/(1/4)
a = 12.4/1
a = 48

S5 = a.(1-r^n)/(1-r)
S5 = 48.(1-(1/4)^5)/(1-1/4)
S5 = 48.(1-1/1024)/(3/4)
S5 = 48.(1023/1024).4/3
S5 = 1023/16 = 63 15/16