Question

Suku ke 2 dan suku ke 6 dari suatu deret geometri berturut turut adalah 6 dan 96. Jumlah tujuh suku pertama dari deret tersebut saka dengan.....

Answer 1

Verified answer

Kelas 8 Matematika
Bab Barisan dan Deret Bilangan

^ = pangkat

r = ^(6 - 2)√(U6/U2)
r = ^4√(96/6)
r = ^4√16
r = 2 atau r = -2

Untuk r = 2

a = U2 : r
a = 6 : 2
a = 3

Sn = a . (r^n - 1)/(r - 1)
S7 = 3 . (2^7 - 1)/(2 - 1)
S7 = 3 . (128 - 1)/1
S7 = 3 . 127
S7 = 381 


Untuk r = -2

a = U2 : r
a = 6 : (-2)
a = -3

Sn = a . (r^n - 1)/(r - 1)
S7 = (-3) . ((-2)^7 - 1) / (-2 - 1)
S7 = (-3) . (-128 - 1) / (-3)
S7 = -129

Answer 2

Verified answer

DeReT GEOmetri

U2 = ar = 6
U6 = ar^5 = 96

U6/U2 = 96/6
r^5 /r = 16
r^4 = 2^4 atau r^4 = (-2)^4

r = 2 atau r = -2

r = 2
ar = 6
a = 6/2 = 3
S7 = a(r^7 - 1)/(r - 1)
S7 = 3(2^7 - 1) / (2 - 1)
S7 = 3(128 - 1)
S7 = 3 × 127
S7 = 381

Utk r = -2
ar = 6
a = 6/-2 = -3

S7 = a(r^7 - 1)/(r - 1)
S7 = -3((-2)^7 - 1) / (-2 - 1)
S7 = -128 - 1
S7 = -129