Suatu barisan aritmatika diketahui U2+U5=1 dan U4+U5=21. Tentukan: A.suku ke-15 B.jumlah 12 suku pertama

U4 + U5 = 21. ==» (a+3b) + (a+4b) = 21
2a + 7b = 21....1)
U2 + U5 = 1. ==» (a+b) +(a + 4b) = 1
2a + 5b = 1. ...2) eliminasi..1) dan ...2)
2a + 7b = 21
----------------- -
-2b = -20
b = -20/-2 = 10
dari..2) 2a + 5b = 1
2a + 5.10 = 1
2a + 50 = 1
' 2a = 1 -50
' 2a = -49
a = -49/2
U15 = a + 14b
= -49/2 + 10
= -49/2 + 20/2
= -29/2
= -14 1/2
Sn = n/2 { 2a + (n-1)b}
S15 = 15/2 { 2.(-49/2) + 14.10 }
= 15/2 { -49 + 140}
= 15/2 .91
= 682 1/2