dik barisan geometri dengan u3=4/9 dan u6=4/243. dit suku awal ?

[tex]U_n=ar^{n-1} \\ U_3= \frac{4}{9} -> ar^2=\frac{4}{9} \\ U_6=\frac{4}{243}->ar^5=\frac{4}{243} \\ \frac{ar^5}{ar^2}= \frac{\frac{4}{243}}{\frac{4}{9}} \\ r^{5-2}=\frac{4}{243}.\frac{9}{4} \\ r^3= \frac{1}{27} \\ r=\frac{1}{3} \\ ar^2=\frac{4}{9} \\ a.(\frac{1}{3} )^2=\frac{4}{9} \\ a.\frac{1}{9}=\frac{4}{9} \\ a=\frac{4}{9}.9 \\ a=4[/tex]
jadi suku pertamanya adlh 4

semoga membantu ya ^_^