Ask your own question, for FREE!
Mathematics 78 Online
OpenStudy (anonymous):

Given the geometric sequence where a1 = 3 and the common ratio is -1, what is the domain for n? a. All integers where ≤ 0 b. All integers where n ≥ 0 c. All integers where n ≤ 1 d. All integers where n ≥ 1 Would the answer be A considering that the ration is a negative?

jimthompson5910 (jim_thompson5910):

The domain of ANY sequence is ALWAYS the set of natural numbers (aka the counting numbers). So the set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...} So that's why the answer is d. All integers where n ≥ 1

jimthompson5910 (jim_thompson5910):

The fact that the common ratio is negative doesn't change that above fact.

OpenStudy (anonymous):

So would the answer for these types of questions always be n ≥ 1?

OpenStudy (anonymous):

And would it not be n ≥ 0 because 0 isn't a natural number?

jimthompson5910 (jim_thompson5910):

yes the domain is always "All integers where n ≥ 1 " why? because when you refer to any term, you're referring to natural numbered terms. You can't have the negative 5th term or the 0th term. So that's why you can only use positive whole numbers.

jimthompson5910 (jim_thompson5910):

no 0 is NOT included

OpenStudy (anonymous):

Okay, thank you!

jimthompson5910 (jim_thompson5910):

oh nvm, you said it differently and I misread, yes 0 is not a natural number or a counting number, so n ≥ 0 is not the answer

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (anonymous):

Ty SM!! This was correct, and indeed was applicable for a lot of the other probs.. :D

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!