suatu deret aritmatika mempunyai suku ke 5 bernilai 30 dan suku ke 12 bernilai 16.jumlah tiga pulu jb suku pertama deret tersebut adalah......

Dik : U12 = a + 11b = 16
U5 = a + 4b = 30
Dit : S30?
Jawab :
Pakai cara eliminasi (untuk mencari b)
a + 11b = 16
a + 4b = 30
—————— -
7b = -14
b = -14/7
b = -2
Untuk mencari a, substitusikan :
a + 4b = 30
a + 4(-2) = 30
a + (-8) = 30
a = 30 + 8
a = 38

Sn = n/2 (2a + (n-1) b)
S30 = 30/2 (2.38 + (30-1)(-2))
= 15 (76 + (29)(-2)
= 15 (76 + (-58)
= 15 (76-58)
= 15(18)
= 270

u5 = 30
a+(n-1)b=30
a+4b=30
a = 30-4b
u12 = 16
a+11b = 16
30-4b+11b = 16
7b = 16-30
b =- 14/7
b = -2

a = 30 - 4(-2)
a = 38

s30 = 30/2 (2x38+(30-1)(-2)
s30 = 15 (76-58)
s30 = 15 x 18
s30 = 270