Solving for n in Geometric Progression

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Homework Statement


The first 3 of a geometric progression are as follows:
[tex]2^{n}, 2^{n+1}, 2^{n+2}[/tex]
Find n

Homework Equations


For this kind of question, the only possibly helpful equation I can think of would be:
[tex]T_{1}=a[/tex]

The Attempt at a Solution


The problem I have is that at first glance I thought I couldn't find the answer to this question because there is not enough information. I know the rate of the progression and can prove it, but it seems physically impossible to find n. The answer in the book is [tex]n=\frac{1}{2}[/tex]
Am I right to assume I cannot find n or is there something I am missing? (Besides more information for the question).
 
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I don't understand why n must be a whole or natural number.
Also, the answer suggests I can find n to be 1/2; is this possible?
 
How does the book define "geometric progression"?
 
Something along the lines of 'a sequence where each progressing term is in the same constant multiple ratio throughout'

e.g. T1,T2,T3...Tn-1,Tn

In this sequence, for it to be a geometric progression: [tex]\frac{T_{2}}{T_{1}}=\frac{T_{3}}{T_{2}}=\frac{T_{n}}{T_{n-1}}[/tex]
 
Hmm...then for any value of n, 2n+1/2n = 2n+2/2n+1 = 2. It seems to me that the problem is bogus.
 
Yes I've been trying out a few things, for any real value of n, the sequence is geometric and the 'rate' of the multiple is 2. There must be a typo in the question somewhere I guess...
 
2 is the ratio multiple of the geometric sequence, not the necessary value for the exponent, n. It can also easily be seen what the ratio is, simply because each progressing term has 1 more in the exponent on the base 2.
On another note: I just used the answer they gave me as n=0.5 to attempt the next sub-question. My answer and the book's answer were completely different; so I suspect it was a typo in the question.