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subst. b=3 ke persamaan U7 (bsa jg persmaan U11)
U7 = a + 6b = 16
<=> a + 6(3) = 16
<=> a + 18 = 16
<=> a = -2
U40 = a + 39b = -2 + 39(3) = -2 + 117 = 115
U7 = a + (7-1) b
16 = a + 6b
U11 = a + (11-1)b
28 = a + 10b
Sekarang di eliminasi :
16 = a + 6b
28 = a + 10b -
-12 = -4b
b = 12/4
b = 3
Sekarang disubstitusikan :
16 = a + 6b
16 = a + 6(3)
16 = a + 18
16-18 = a
-2 = a
Un = a + (n-1)b
U40 = -2 + (40-1)3
U40 = -2 + (39)3
U40 = -2 + 117
U40 = 115
Semoga membantu dan maaf kalo salah :))