Pliss yg pinter tolong jawab kan

A.) Sn = 1/2n (2a + (n-1) b) = 17

➡ S4 = 1/2(4) (2a + (4-1) b) = 17
S4 = 2 ( 2a + 3b ) = 17
S4 = 4a + 7b = 17

➡ S8 = 1/2(8) (2a + (8-1) b) = 58
S8 = 4 ( 2a + 7b ) = 58
S8 = 8a + 28b = 58

Eliminasi
S4 = 4a + 7b = 17 (dikali 2)
S8 = 8a +28b = 58 (dikali 1)
= 8a + 14b = 34
= 8a + 28b = 58
____________-
= - 14b = -28
b = 2
Substitusi
4a + 7b = 17
4a + 7(2) = 17
4a + 14 = 17
4a = 17 - 14
a = 3/4




B.) U9 = a + (n-1) b
= 3/4 + (9-1) 2
= 3/4 + 16
= 3/4 + 16/1
= 3/4 + 32/4
= 35/4

C.) S9 = 1/2n (2a + (n-1) b)
= 1/2(9) (2x3/4 + (9-1) 2)
= 1/2 x 9 ( 6/4 + 16 )
= 1/2 x 9 ( 6/4 + 16/1)
= 1/2 x 9 ( 6/4 + 64/4)
= 1/2 x 9 (70/4)
= 70/8 x 9
= 78,75


Semoga membantu :)