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U2= a + 1b = 127
a + 1b = 127 (persamaan 1)
u4= a + 3b = 3
a+ 3b= 3 (persamaan 2)
~eliminasi persamaan2 ter sebut
a + 1b =127
a + 3b = 3 -
-2b = 124
b= 124/3
b= 40 4/3
mencari a
a + 3b = 3
a + 3(40 4/3) = 3
a= 9 - 160
a= - 151
maka
U9= a + (n-1)b
u9= -151 + (8 x 40 4/3)
U9= -151 + 426,67
u9= 275, 67
U₂ = ar
3 = ar³
129 = ar
r² =
r = 1/7
a= 129 : 1/7 = 129 × 7 = 903
U₉ = 903 . (1/7)⁸
= 903 .
= 1,566
saya rasa ada kesalahan soal.