Question 928744
By definition, 


a2 = r*a1


a3 = r*a2


a3 = r*(r*a1)


a3 = r^2*a1


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Since a3 = r^2*a1, we know 



a3 = r^2*a1


16 = r^2*2


16/2 = r^2


8 = r^2


r^2 = 8


r = sqrt(8)


r = 2*sqrt(2)


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The general nth term formula is 



an = a1*(r)^(n-1)


an = 2*(2*sqrt(2))^(n-1)



Now plug in n = 6



an = 2*(2*sqrt(2))^(n-1)


a6 = 2*(2*sqrt(2))^(6-1)


a6 = 2*(2*sqrt(2))^5


a6 = 2*(128*sqrt(2))


a6 = 256*sqrt(2)


So {{{a6 = 256*sqrt(2)}}}

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