Suatu barisan aritmetika mempunyai U6=23 dan U10=43.Jumlah lima suku pertama barisan tersebut adalah

b = 43-23/10-6 = 20/4 = 5

U6 = a + 5b
23 = a + 5(5)
a = 23-25 = -2

Maka

S5 = 5/2 (2(-2) + 4(5))
= 5/2(-4+20)
= 5/2 (16)
= 40

Materi : Barisan dan deret
Kelas : IX SMP

U6 = U1 + 5b = 23
U10 = U1 + 9b = 43
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========> -4b = -20
========> b = 5

U6 = U1 + 5b
23 = U1 + 5(5)
23 = U1 + 25
U1 = 23 - 25
U1 = -2

Sn = n/2 ( 2.U1 + (n - 1) b)
S5 = 5/2 ( 2(-2) + ( 5 - 1) 5)
S5 = 2,5 ( -4 + 4 . 5)
S5 = 2,5 (-4 + 20 )
S5 = 2,5 ( 16)
S6 = 40