SOLUTION: Find a8 of a geometric sequence given that a3 = 32 and a5 = 512.

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Question 192065This question is from textbook Saxon Algebra 2
: Find a8 of a geometric sequence given that a3 = 32 and a5 = 512. This question is from textbook Saxon Algebra 2

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
You can review this site:
http://www.regentsprep.org/Regents/math/algtrig/ATP2/GeoSeq.htm
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An = A1(r^(n-1))
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The idea is that because they gave you:
a3 = 32 and a5 = 512
you can now find A1(first term) and r (the common ratio).
Now, you can find any term after that.
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An = A1(r^(n-1))
32 = A1(r^2)
A1 = 32/r^2
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An = A1(r^(n-1))
An = A1(r^4)
512 = (32/r^2)(r^4)
512 = 32(r^2)
16 = r^2
4 = r (common ratio)
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A1 = 32/r^2 = 32/4^2 = 32/16 = 2 (first term)
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An = A1(r^(n-1))
a8 = (2)(4^(8-1))
a8 = (2)(4^7)
a8 = 2*16384
a8 = 32768