6. Suatu deret aritmatika dengan a = -18, b = 3. Jika Sn = 0, hitunglah : a. Besar suku akhir (Un)! b. Banyaknya suku deret tersebut ! c. Tulislah deret aritmat

Sn = n/2 ( 2(-18) + (n-1)3)
0 = n(-36 + 3n -3)
0 = 3n² - 39n
3n²-39n = 0
n² - 13n = 0
n(n-13) = 0
n = 0 (TM)
atau
n= 13

a. Un = U13
U13 = a+ 12b
= -18 + 12(3)
= 36 - 18
= 18

b.Banyak suku(n) = 13


c.Sn= S13
S13 = 13/2 (2(-18) + 12(3)
= 13/2 (-36 + 36)
= 13/2 (0)
= 0 (Terbukti)

S13 = -18 -15-12-9-6-3-0-3+6+9+12+15+18 =0