SOLUTION: 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.

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)

-------------------------------------------------------

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): You can put this solution on YOUR website!
since you have a_1, the next term is a_1*r, third term is a_1*r^2.

<-- we don't know which one.
Then

However the value of a6 depends on what r is. If then . Otherwise . Impossible to determine what a_6 is without more info.
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