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((1)/(4)\right)=\boxed{(3)/(4)\cdot(-4)^n}\\\\Answer:\boxed{a_n=3\cdot(-4)^(n-1)=(3)/(4)\left(-4\right)^n}`