Question

3,-12,48,-192 sequence

Answer 1

$a_1=3;\ a_2=-12;\ a_3=48;\ a_4=-192\\\\a_2=-4a_1\to a_2=-4(3)=-12=a_2\\\\a_3=-4a_2\to a_3=-4(-12)=48=a_3\\\\a_4=-4a_3\to a_4=-4(48)=-192=a_4\\\\It's\ a\ geometric\ sequence\ where\ a_1=3\ and\ r=-4.$

$The\ n-th\ term\ of\ a\ geometric\ sequence\ with\ initial\ value\\a\ and\ common\ ratio\ r\ is\ given\ by:\\\\a_n=a_1r^{n-1}\\\\subtitute\\\\a_n=3\cdot(-4)^{n-1}=3\cdot(-4)^n\cdot(-4)^{-1}=3\cdot(-4)^n\cdot\left(\frac{1}{4}\right)=\boxed{\frac{3}{4}\cdot(-4)^n}\\\\Answer:\boxed{a_n=3\cdot(-4)^{n-1}=\frac{3}{4}\left(-4\right)^n}$