Geometric Sequence: T1=0.1024, T2=0.256, Middle Term=156.25

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The discussion revolves around determining the number of terms in a geometric sequence where T1 is 0.1024, T2 is 0.256, and the middle term is 156.25. The initial calculation suggests there are 18 terms based on the formula for the middle term. However, the conversation reveals a critical misunderstanding about the definition of a middle term in sequences with an even number of terms. It is clarified that a sequence can only have a middle term if the total number of terms is odd, leading to the conclusion that there should actually be 17 terms in this sequence. The participants emphasize the importance of correctly identifying the relationship between the number of terms and the middle term's position.
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A finite geometric sequence has t1 = 0.1024 and t2 = 0.256. How many terms does this sequence have if its middle term has a value of 156.25?

My Solution

Common Ratio: T2/T1=(.256)/(.1024)=2.5
What term # is the middle term?
tn=ar^n-1
a=0.1024
r=2.5
tn=156.25

(156.25)=(0.1024)(2.5)^n-1

1525=(2.5)^n-1

[Log(1525)/Log(2.5)]+1=n
n=8+1=9

n is the middle term so final term should be 2n. 18 terms in the sequence

From the way I have seen other people do this question, they get the answer "17 terms". Why am I 1 term off? Can you help me with what I am doing wrong. Or am I doing it correctly :)

Thanks!
 
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ThomasMagnus said:
n is the middle term so final term should be 2n. 18 terms in the sequence
Not so! Can a sequence with an even number of terms have a middle term?

For example, if I have a simple sequence of natural numbers where 3 is the 3rd and middle term:
1 2 3 ...
You're saying that there are 2n, or 6 terms in the sequence:
1 2 3 4 5 6
See the problem? If 3 is the middle term, then there are 2n - 1, or 5 terms in the sequence:
1 2 3 4 5

So there should be 17 terms in the sequence in your problem.
 
eumyang said:
Not so! Can a sequence with an even number of terms have a middle term?

For example, if I have a simple sequence of natural numbers where 3 is the 3rd and middle term:
1 2 3 ...
You're saying that there are 2n, or 6 terms in the sequence:
1 2 3 4 5 6
See the problem? If 3 is the middle term, then there are 2n - 1, or 5 terms in the sequence:
1 2 3 4 5

So there should be 17 terms in the sequence in your problem.
I'm struggling to understand what you are saying.

"You're saying that there are 2n, or 6 terms in the sequence:
1 2 3 4 5 6"

Isn't there six terms in this sequence?Thanks :)
 
ThomasMagnus said:
I'm struggling to understand what you are saying.

"You're saying that there are 2n, or 6 terms in the sequence:
1 2 3 4 5 6"

Isn't there six terms in this sequence?
Yes, but I consider this sequence to have no "middle term", because there are an even number of terms. If you have a 5 term sequence (2n - 1):
1 2 3 4 5
... then 3 would be the middle term.
 
So the middle term is always defined as 2n-1?
 
Yes, if there are n terms in a sequence and n is odd.
 
ThomasMagnus said:
So the middle term is always defined as 2n-1?

eumyang said:
Yes, if there are n terms in a sequence and n is odd.
No, if there are n terms, for n larger than 1, then 2n- 1 is larger than n! If n is odd, then the middle term is indexed by (n+ 1)/2.
 
HallsofIvy said:
No, if there are n terms, for n larger than 1, then 2n- 1 is larger than n! If n is odd, then the middle term is indexed by (n+ 1)/2.

Wow. I need to wake up. I think I meant to say: if the middle term of a sequence is the nth term, then there are 2n - 1 terms in the sequence. :blushing:
 
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