Question

Diketahui barisan aritmetika dengan U2=12 dan U5+U6=94,maka jumlah 12 suku pertama pada barisan tersebut adalah

Answer 1

U2 = 12
U5 + U6 = 94
S12 = ?

U2 = a + b = 12 ........................................(pers. 1)
U5 + U6 = (a + 4b) + (a + 5b) = 94
                   2a + 9b = 94...........................(pers. 2)

a + b = 12       |2|
2a + 9b = 94   |1|

2a + 2b = 24
2a + 9b = 94
__________ _
        -7b = -70
           b = 10

a + b = 12
a + 10 = 12
a = 2

S12 = n/2 {2a + (n-1)b}
       = 12/2 (2.2 + 11.10)
       = 6 ( 4 + 110)
       = 6 (114)
       = 684




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Answer 2

U2 = a + b = 12<br />U5 + U6 = a +4b + a + 5b<br /> 94 = 2a +9b<br />Eliminasikan<br />2a + 9b = 94 (dikalikan 1)<br />a + b = 12 (dikalikan 2)<br />maka:<br />2a +9b = 94<br />2a +2b = 24<br />dikurangkan<br />= 7b = 70<br />b = 10<br />nilai a = a + b = 12<br />a + 10 = 12<br />a = 2<br />S12 = n/2. (2a +( n - 1)b)<br /> = 12/2 ( 2.2 + ( 12 -1 ) 10<br /> = 6 (4 +110) = 6 ( 114) = 684