How do you find A1 in a geometric sequence if A3=-8 and A6=1?

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Answer 1 · 2015-11-08T10:07:10

#A_1 = -32#

#A_n = A_1r^(n -1)#

#A_3 = -8 = A_1r^(3-1)#

#=> -8 = A1r^2#

#A_6 = 1 = A_1r^(6-1)#

#=> 1 = A_1r^5#

#=> 1 = A_1r^2r^3#

But #-8 = A1r^2#

#=> 1 = -8r^3#

#=> -1/8 = r^3#

#=> r = -1/2#

#A_3 = A_1r^(3-1)#

#=> -8 = A_1*(-1/2)^2#

#=> -8 = A_1*1/4#

#=> -32 = A_1#