dalam suatu deret aritmatika suku ke 4 = 23 dan bedanya = 5 . tentukan jumlah 12 suku pertamanya

U4 = 23
b = 5

Un = a+(n-1)b
U4 = a+(4-1)b
23 = a+3(5)
23 = a+15
a = 23-15
a = 8

Sn = n/2 (2a + (n-1)b)
S12 = 12/2 (2(8) + (12 - 1)5)
S12 = 6(16 + 55)
S12 = 6(71)
S12 = 426

U4 = a + 3b
23 = a + 3(5)
23 - 15 = a
8 = a

S12
= n/2 {2a + (n-1)b}
= 12/2 {2(8) + 11(5)}
= 6 (16 + 55)
= 6 (71)
= 426