jika suku-suku 2k+2,K+7,2k+2,k+7 dan 3k+63k+6 merupakan tiga suku pertama berurutan barisan aritmatika, tentukan besarnya suku ke 11 ?

Barisan Aritmatika

(2k + 2) , (k + 7) , (3k + 6)

U2 - U1 = U3 - U2
(k + 7) - (2k + 2) = (3k + 6) - (k + 7)
-k + 5 = 2k - 1
2k + k = 5 + 1
3k = 6
k = 2

a = 2k + 2 = 2.2 + 2
a = 6

b = U2 - U1 = -k + 5
b = -2 + 5
b = 3

U11 = a + 10b
U11 = 6 + 10 . 3
U11 = 36

2 U2 = U1 + U3
2 (k+7) = 2k+2 + 3k+6
2k + 14 = 5k + 8
14-8 = 5k-2k
3k = 6
k = 6/3 = 2

A= U1 = 2(2)+2 = 6
U2 = k+7 = 2+7 = 9
B = U2-U1 = 9-6= 3

Un = a+(n-1)b
U11 = 6+(11-1)3
U11 = 6+ 30 = 36