A geometric series #a_n# starts from a certain number #a_1=x# and computes the next term #a_{n+1}# by multiplying the previous one by a fixed ratio #r#. So, you have
#a_1=x#
#a_2=a_1*r=x*r#
#a_3 = a_2*r = (x*r)*r=x*r^2#
#a_4 = a_3*r = (x*r^2)*r=x*r^3#
So, it should be clear that the general rule is
#a_n = x*r^{n-1}#
So, to find the generi expression, we need both #x# and #r#. Note that, since we have the first term in each series, we immediately know #x#: you just need to look at the first term, since #a_1=x#.
So, if we call the first series #a_n# and the second #b_n#, we have #a_1=4# and #b_1=16#.
To compute #r#, we can divide two consecutive terms, since you have
#a_{n+1}/a_n = (x*r^n)/(x*r^{n-1})=r#
So, if we call #r_a# the ratio of the first series and #r_b# the ratio of the second series, and we use the first two terms to compute it, we have
#r_a = a_2/a_1=-4/4=-1,\qquad r_b=b_2/b_1=8/16=1/2#
So, we have the general rules:
#a_n = 4*(-1)^{n-1},\qquad b_n = 16*(1/2)^{n-1} = 2^4*2^{-n+1} = 2^{5-n}#
To compute the eight element, simply plug #n=8#:
#a_8 = 4*(-1)^7 = 4*(-1) = -4#
#b_8 = 2^{5-8} = 2^{-3} = 1/8#
