Question 928744: Hello!
I am having a problem on a question involving geometric sequences and series. The problem gives me a1= 2 and a3=16 and wants to know a6=?. I have tried plugging in r=2, 2.5, 3, 3.5, 4, 4.5 (these haven't worked) so I am wondering if it is a radical of some kind.? Any help would be greatly appreciated! Thank you!
K. Legters
Found 2 solutions by jim_thompson5910, richard1234:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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 
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Answer by richard1234(7193)  (Show Source):
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