Question

8th term of 1/4,1/8,1/16,1/32

Answer 1

$a_1=\dfrac{1}{4};\ a_2=\dfrac{1}{8};\ a_3=\dfrac{1}{16};\ a_4=\dfrac{1}{32};...\\\\it's\ a\ geometric\ sequence\ where\ a_1=\dfrac{1}{4}\ and\ r=\dfrac{1}{8}:\dfrac{1}{4}=\dfrac{1}{2}\\\\a_n=a_1r^{n-1}\\\\n=8\\\\a_8=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)^{7-1}=\dfrac{1}{4}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{4}\cdot\dfrac{1}{64}=\dfrac{1}{256}$