Jumlah 13 suku pertama dari deret aritmetika (k + 1),2k,(4k - 6) adalah

suku awal a = k+1 dan beda b = 2k -(k+1) = (4k -6) - (2k) = k-1 = 2k-6
k = 5
Jadi deret itu : 6, 10, 14,
Sn = n/2 { 2a + (n-1)b }
S13 = 13/2 ( 2.6 + 12.5)
S13 = 13/2 (12 + 60)
S13 = 13/2 × 72
S13 = 13× 36
S13 = 468

jawab

sifat aritemetika = u1 + u3= 2 u2
k+ 1  +k - 6 = 2(2k)

2k - 6 = 4k
-2k = 6
k  = - 3

a = k + 1
a = - 3 + 1
a = -2

b = 2k - k - 1
b = k  -1
b = -3-1
b = - 4

Sn = n/2 ( 2a + (n-1)b))
s13 = 13/2 (2(-2)+ 12(-4))
s13 = 13/2 (- 4 - 48)
s13 = 13/2 (-52) = - 338