Deret aritmatika U3 = 6 dan U8 = -4 jumlah 12 suku pertamanya?

Dik. Barisan deret aritmatika :
u3 = 6 & u8 = -4.
Dit. S12 ?

Jawab...

u3 = a + 2b = 6
u8 = a + 7b = -4
      ----------------------- (-)
-5b = 10
b = -2

a :
a + 2b = 6
a + (2) (-2) = 6
a - 4 = 6
a = 6 + 4
a = 10

Sn = n/2 ( 2a + (n - 1) b)
S12 = 12/2 (2(10) + (11) (-2))
S12 = 6 (20 - 22)
S12 = 6 (-2)
S12 = -12 

Semoga membantu... ^_^