Question

Diketahui jumlah 3 suku pertama dari deret geometri adalah 117 dan jumlah 3 suku berikutnya adalah 13/3. tentukan suku ke 5 dari deret tersebut?

Answer 1

 u1+u2+u3 = 117 --> a + ar + ar^2 = 117
a(1 + r + r^2) = 117

 u4 +u5+u6 =13/3--> ar^3 + ar^4 + a^5 = 13/3
ar^3( 1 + r+r^2) = 13/3
..
ar^3 ( 1+ r+ r^2) / a (1+r+r^2) = 117/ (13/3)
r^3 =27
r = 3
.
a (1+r+r^2) = 13/3
a (1 +3 + 9) = 13/3
13 a = 13/3
a = 1/3
..
U5 = ar^4  = 1/3 . 3^4 = 3^3 = 27 

Answer 2

S3 = a ( 1 - r^3) / ( 1 -r) = 117

S6 = a ( 1 -r^6) / ( 1-r) = 13/3 + 117 = 364/3

S6/S3 =   (364/3) : 117 = 364/ 351

S6 / S3 = 364/351

(1 - r^6) / ( 1 - r^3) = 364 / 351
( 1 + r^3 ) ( 1-r^3) / (1-r^3) = 364 /351
1 + r^3 = 364 /351
351 + 351r^3 = 364
351r^3 = 364 - 351
r^3 = 13/351= 1/27
r = 1/3

a( 1- r^3) / (1-r) = 117

a ( 26/27) / ( 2/3) = 117
a ( 13/9 ) = 117
a = 117 x 9/13 = 81

u5 = a r^4 = 81 (1/3)^4 =81/81 = 1