What is the 12th term of a geometric sequence where #a_1 = 8# and #a_6 = -8,192#?

Let K be a constant.
Let the geometric ratio be r.
Let #a_n# be the resulting #n^(th) color(white)(x) #term of the sequence

Then at

#n=1# we have #a_n= kr^n = (+8)#
#n=6# we have #a_n=kr^n=(-8192)#

The fact that #a_6# is negative implies a switching between positive and negative sequence. This is achieved by the incorporation of #(-1)^(n+1)#. It has to be #n+1# or an appropriate variation to force #a_n# to be positive and negative in the appropriate place.

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To find r

Known:
#a_1=(-1)^(1+1)(k)(r^(1)) = (+8) " "-> " "kr=8# ........(1)
#a_6 = (-1)^(6+1)(k)(r^(6)) = -8192" -> " " -kr^(6) = -8192#..(2)

To isolate r use #(2) divide (1)# giving:

#(-kr^6)/(kr) = (-8192)/8#

Thus #r^5 =1025# Note the reversal from negative to positive.

so #r = root(5)(1025) =4#

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To find k by substitution into (1)
#(-1)^2(k)(4)=8#
k=2

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Thus:
#a_12 = (-1)^(12+1) (2)(4)^12= - 33554432#

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