\((1;\ 4)\to a_1=4\\(2;\ 0.8)\to a_2=0.8\\(3;\ 0.16)\to a_3=0.16\\\\it's\ a\ geometric\ sequence\ because\ a_1a_3=a_2^2\\\\a_1a_3=4\cdot0.16=0.64\\\\a_2^2=0.8^2=0.64\)
\(a_n=a_1r^{n-1};\ r=a_2:a_1\to r=0.8:4=0.2=\dfrac{1}{5}\\\\subtitute:\\\\a_n=4\cdot\left(\dfrac{1}{5}\right)^{n-1}\)
