Solving for n in Geometric Progression

Click For Summary

Homework Help Overview

The problem involves finding the value of n in a geometric progression defined by the terms 2^{n}, 2^{n+1}, and 2^{n+2}. Participants discuss the implications of the problem statement and the nature of geometric progressions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the definition of geometric progression and question the necessity of n being a whole or natural number. Some express confusion about the ability to determine n given the terms provided.

Discussion Status

The discussion is ongoing, with participants sharing different interpretations of the problem. Some suggest that the problem may contain a typo, while others explore the implications of various values for n. There is no explicit consensus on how to proceed.

Contextual Notes

Participants note that the problem may lack sufficient information or clarity, leading to differing interpretations of the requirements for n. The mention of a potential typo in the question adds to the uncertainty.

Mentallic
Homework Helper
Messages
3,802
Reaction score
95

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).
 
Physics news on Phys.org
For #1, those are general terms of a geometric sequence. The domain for n is either the whole numbers or the natural numbers. Either domain will give you values for the terms.
 
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...
 
Or, as e(ho0n3 said, in the answer. n= 2 satisfies the problem nicely.
 
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
2K
Replies
4
Views
2K