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17 = U1 + (3-1) . b
17 = U1 + 2b ... (1)
U5 = U1 + (n-1) . b
31 = U1 + ( 5-1) . b
31 = U1 + 4b ... (2)
~~~~~~~~~
17 = U1 + 2b
31 = U1 + 4b
_________-
-4 = -2b
b= -4 : -2
b= 2
sub. b = 2 per (1)
17 = U1 + 2b
17 = U1 + 2 (2)
17 = U1 + 4
17 -4 = U1
U1 = 13
U5=31=a+4b
metode eliminasi
a+4b=31
a+2b=17
2b=14
b=7
a+2.7=17
a=17-14
a=3
U20= 3+(19.7)=136