Pada suatu deret aritmetika,diketahui suku ke-2=5 dan suku ke-4=11,maka u12 adalah
Sakhara14
U2 = 5 U4 = 11 U2 = a + b = 5 U4 = a + 3b = 11 - -2b = -6 b = 3 U2 = a + b = 5 a + 3 = 5 a = 5-3 a = 2
U12 = a + 11b = 2 + (11.3) = 2 + 33 = 35
3 votes Thanks 6
Dzaky111
suku ke 4 ⇒⇒ a + 3b = 11 suku ke 2 ⇒⇒ a + b = 5 _ 2b = 6 b = 3 suku ke 2 ⇒⇒ a + b = 5 a + 3 = 5 a = 2 U12 = a +(n-1)b = 2 + (12-1)3 = 2 + 11.3 = 2 + 33 = 35
U4 = 11
U2 = a + b = 5
U4 = a + 3b = 11 -
-2b = -6
b = 3
U2 = a + b = 5
a + 3 = 5
a = 5-3
a = 2
U12 = a + 11b = 2 + (11.3) = 2 + 33 = 35
suku ke 2 ⇒⇒ a + b = 5 _
2b = 6
b = 3
suku ke 2 ⇒⇒ a + b = 5
a + 3 = 5
a = 2
U12 = a +(n-1)b
= 2 + (12-1)3
= 2 + 11.3
= 2 + 33
= 35