Question

Jika Un adalah suku ke-n suatu deret aritmatika U1+U2+U3=-9 dan U3+U4+U5=15, tentukan jumlah lima suku pertama deret tersebut

Answer 1

U1 + U2 + U3 = -9
a + (a + b) + (a + 2b) = -9
3a + 3b = -9

U3 + U4 + U5 = 15
(a + 2b) + (a + 3b) + (a + 4b) = 15
3a + 9b = 15

Eliminasi:
3a + 3b = -9
3a + 9b = 15
-------------------- (-)
-6b = -24
b = -24/-6
b = 4

Subtitusi:
3a + 3b = -9
3a + 3(4) = -9
3a + 12 = -9
3a = -9 - 12
3a = -21
a = -21/3
a = -7

Jumlah 5 suku pertama:
Sn = n/2 [2a + (n-1)b]
S5 = 5/2 [2(-7) + (5-1)4]
S5 = 2,5 [-14 + 20 - 4]
S5 = 2,5 [2]
S5 = 5

Kalo Jawabannya tepat, aku minta bintang 5 :p

Answer 2

U3+U4+U5=15
U1+U2+U3=-9
---------------------- -
(U4 - U1) + (U5 - U2) = 24
3b + 3b = 24
6b = 24
b = 4

S3 = 3/2(2a + (3-1)b)
-9 = 3(a + b)
-9 = 3a + 3b
-9 = 3a + 12
3a = -21
a = -7

S5 = 5/2(2a + (5-1)b)
S5 = 5(a + 2b)
S5 = 5(-7 + 8)
S5 = 5

CMIIW :3