diketahui deret aritmatika dengan suku ke-3 = 24 dan suku ke-6 = 35.hitunglah jumlah 15 suku pertama deret tersebut?

a + 5b = 35
a + 2b = 24
_________ _
3b = 11
b = 11/3

Substitusi b = 11/3
a + 5b = 35
a + 55/3= 35
a = 50/3

S15 = 15/2 (2×50/3 + 14×11/3)
S15 = 15/2 (100/3+154/3)
S15 = 15/2 (254/3)
S15 = 635

Jumlah 15 suku pertama adalah 635

Dik:
U3 = 24
a + 2b = 24 ---------->> (1)

U6 = 35
a + 5b = 35 ---------->> (2)

Dit:
S15 = ??

Jwb:
Eliminasikan 1 dan 2,,
a + 5b = 35
a + 2b = 24
___________ -
3b = 11
b = 11/3

Masukkan nilai "b" ke 1,,
a + 2b = 24
a + 2(11/3) = 24
a + 22/3 = 24
a = 24 - 22/3
a = 50/3

Jadi,
Sn = n/2(2a + (n-1)b)
S15 = 15/2(2.50/3 + (15-1)11/3)
S15 = 15/2(100/3 + 154/3)
S15 = 15/2(254/3)
S15 = 3810/6
S15 = 635