A puzzling math geometric sequence question.

[barryj]

Homework Statement

This problem is taken directly out of a textbook.

"The first three terms of a geometric sequence are 1,2, and 4. Susanna said the 8th term of this sequence is 128. Paul said the 8th term is 29. Explain how the students found their answers. Why could these both be considered correct answers?

Homework Equations

a(n) = a(0)R^(n-1) terms for a geometric sequence with r being the common ratio, a(0) being the first term.

The Attempt at a Solution

Obviously, a(8) = 1(2)^(8-1) = 128.

So how could 29 be a correct answer?

The teachers edition of the algebra textbook does not give the solution to why 29 could be a correct answer.

 

[Ray Vickson]

Homework Statement

This problem is taken directly out of a textbook.

"The first three terms of a geometric sequence are 1,2, and 4. Susanna said the 8th term of this sequence is 128. Paul said the 8th term is 29. Explain how the students found their answers. Why could these both be considered correct answers?

Homework Equations

a(n) = a(0)R^(n-1) terms for a geometric sequence with r being the common ratio, a(0) being the first term.

The Attempt at a Solution

Obviously, a(8) = 1(2)^(8-1) = 128.

So how could 29 be a correct answer?

The teachers edition of the algebra textbook does not give the solution to why 29 could be a correct answer.

What happens if you represent integers in a base other than 10?

 

[barryj]

Maybe, this is a problem out of the Algebra 1 textbook by Holt. I don't think they know about other number bases at that point but perhaps this is the answer. If it is a different base, it would have to be 10 or greater to have a 9 as a digit, yes?

 

[scurty]

a(n) = a(n-1) + n, given a(0) = 1.

Basically, add 1 to the first term to get the second term, add 2 to the second term to get third, etc...

Edit: Just realized that my formula isn't in the form of a geometric sequence.. you can ignore my post.

 

[Dick]

What happens if you represent integers in a base other than 10?

128 is the base 10 answer and it's even. 29 in any integer base is odd, isn't it?

 

[barryj]

Well, one of my students found an answer. The sequence is not geometric.

a(1) = 1
a(n) = a(n-1) + (n-1)

giving the sequence 1,2,4,7,11,16,22,29

 

[Dick]

Well, one of my students found an answer. The sequence is not geometric.

a(1) = 1
a(n) = a(n-1) + (n-1)

giving the sequence 1,2,4,7,11,16,22,29

Yeah, I managed to find a version of this question online here http://www.nbisd.org/users/0006/docs/Textbooks/Algebra1/A1c11.pdf. Note that the original problem doesn't have the word 'geometric' in it.

 

[scurty]

Well, one of my students found an answer. The sequence is not geometric.

a(1) = 1
a(n) = a(n-1) + (n-1)

giving the sequence 1,2,4,7,11,16,22,29

I know I said to ignore my post, but I didn't mean completely. :(

 

[Dick]

I know I said to ignore my post, but I didn't mean completely. :(

Well, I looked at it. It was a good point. Turned out to be correct once 'geometric' was deleted from the problem statement.

 

[barryj]

My textbook, a Holt Algebra 1, did have the word geometric in it. I guess the later version had the problem corrected. It's still an interesting problem.

 

[Dick]

My textbook, a Holt Algebra 1, did have the word geometric in it. I guess the later version had the problem corrected. It's still an interesting problem.

Well, that's sloppy. And more annoying than interesting, I would say. A sequence starting with 1,2,4 can have any 8th term you want, if you are sufficiently creative with the rule that generates it. Check out http://oeis.org. I get 9014 hits on a sequence containing 1,2,4.