SOLUTION: WHAT IS THE 10TH TERM OF THE GEOMETRIC SEQUENCE 64,-32,16,8???

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Question 250626: WHAT IS THE 10TH TERM OF THE GEOMETRIC SEQUENCE 64,-32,16,8???
Found 2 solutions by drk, palanisamy:
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
We have to find a pattern. First, geometric sequences have the following formula:
(I) t%28n%29+=+a%28r%29%5E%28n-1%29
Here are the numbers: 64, -32, 16, 8.
t(1) = 64
t(2) = -32
t(3) = 16
t(4) = -8
Put t(1) into (i) and we get
64 = ar^(1-1)
** 64 = a **.
Put t(2) = -32 into (i)and include a = 64. We get
-32 = 64r^(2-1)
-32 = 64r^1
divide by 64 and we get
** r = -1/2 **.
We know a and r. Now we write the formula as
T%28n%29=+64%28-1%2F2%29%5E%28n-1%29
We want the 10-th term, or n = 10. SO,
t%2810%29+=+64%28-1%2F2%29%5E%2810-1%29
t%2810%29+=+64%282%29%5E%28-9%29 = 64/512 = .125

Answer by palanisamy(496) About Me  (Show Source):
You can put this solution on YOUR website!
The given Geometric sequence is 64,-32,16,-8,4,...
The first term a = 64
The common ratio r = -1/2
The nth tern Tn = ar^(n-1)
Tn = 64*(-1/2)^(n-1)
Put n=10.
The 10th term T10 = 64*(-1/2)^(10-1)
= 64*(-1/2)^9
= -(2^6)/2^9
= -1/2^3
= -1/8