suatu deret aritmetika diketahui u3+u4=5 dan u5+u7=40. jumlah 10 suku pertama dari deret tersebut adalah

Aritmatika

U3 + U4 = 5
a + 2b + a + 3b = 5
2a + 5b = 5

U5 + U7 = 40
a + 4b + a + 6b = 40
2a + 10b = 40

Gunakan metode eliminasi
2a + 5b = 5
2a + 10b = 40
___________-
-5b = -35
b = 7

Gunakan metode subtitusi
2a + 5b = 5
2a = 5 - 5(7)
2a = -30
a = -15

Sn = n/2 (2a + (n - 1)b)
S10 = 10/2 (2.(-15) + 9(7))
S10 = 5 (-30 + 63)
S10 = 5 (33)
S10 = 165

semoga berguna +_+

Arithmetic Series.

U₃ + U₄ = 5
a + 2b + (a + 3b) = 5 → 2a + 5b = 5

U₅ + U₇ = 40
a + 4b + (a + 6b) = 40
2a + 10b = 40 → a + 5b = 20

2a + 5b = 5
a + 5b = 20
---------------- -
a = -15
-15 + 5b = 20 → b = 7

Sn = n/2 [2a + (n - 1)b]
S₁₀ = 10/2 [2(-15) + (10 - 1)(7)] = 165