Question

I understand that the formula is a2 over a1 and so on, but I am not quite sure how to do fractions..
"Determine whether the sequence is geometric. Explain why or why not."
16/3, 8/3, 4/3, 2/3, 1/3

Answer 1

A sequence is geometric if there is a single number that when you multiply a term you get the next term. This must hold for every term.

Answer 2

I know - a common ratio, right?

Answer 3

Exactly.
To go from 16/3 to 8/3, you are dividing by 2. Dividing by 2 is the same as multiplying by 1/2.
Now let's see, if you multiply the first term by 1/2, do you get the second term?

Answer 4

16/3 * 1/2 = 8/3
Yes you do.

Answer 5

I Kind of sort of understand what you are saying..

Answer 6

Now let's try the same between the second and third terms.
Is 8/3 * 1/2 = 4/3
Yes it is, so the same common ratio applies to the second to third terms.

If you keep doing it, you will see that the same factor of 1/2 will give you the next terms. That makes it a geometric sequence.

Answer 7

Yes I could do that, but I also cannot. See I was given a specific way to do this and I am afraid that if I do not, I will not get credit.

Answer 8

Perhaps the way you were told to do it is to divide each term by the previous term, and see if you always get the same result.

Answer 9

Yes that is the exact way I was told to do it, but that does not seem to have worked out because I marked it as not geometric.

Answer 10

Let's do it the way you were taught.

Answer 11

Two divided by five halves would be.. four fifths??

Answer 12

Divide the second term by the first term.

\(\dfrac{8}{3} \div \dfrac{16}{3} \)

\(= \dfrac{8}{3} \times \dfrac{3}{16}\)

\( = \dfrac{8 \times 3}{3 \times 16}\)

\(= \dfrac{8 \times 3}{3 \times 8 \times 2} \)

\(= \dfrac{\cancel{8} \times \cancel{3} 1}{\cancel{3} \times \cancel{8} \times 2} \)

\(= \dfrac{1}{2}\)

Answer 13

Wow. I am an idiot.

Answer 14

As you can see, the second term divided by the first term is 1/2.
Now you need to do the same for the third term divided by the second term, and so on.

Answer 15

I have been looking at the stinking wrong problem this whole time. I do understand now though, thank you. I was wondering if you could help me with a second sequence as well? If not, that is fine.

Answer 16

I don't agree with you. If you were an idiot, you wouldn't be asking questions on OS. Since you want to learn this you are not an idiot. The way to learn is to ask questions.

Answer 17

Sure, let's see the other sequence.

Answer 18

Thank you that is very kind of you to say. Okay here is the second set :
5/2, 2, 3/2, 1, 1/2

Answer 19

Let's do the first division, a2/a1:

\(\dfrac{5}{2} \div 2\)

\(=\dfrac{5}{2} \div \dfrac{2}{1} \)

\(=\dfrac{5}{2} \times \dfrac{1}{2} \)

\(= \dfrac{5}{4} \)

Ok so far?

Answer 20

Yes

Answer 21

Wait no, wouldn't it be 4/5 because 5/2 is a1

Answer 22

Sorry, you are correct. Let me do that again.
I did a1/a2 by mistake.
Wow, I'm impressed. you really are paying attention.

Answer 23

I'm not quite sure if that should be taken as a compliment. But thanks anyways? haha

Answer 24

\(2 \div \dfrac{5}{2}\)

\(= \dfrac{2}{1} \times \dfrac{2}{5} \)

\(= \dfrac{4}{5} \)

Answer 25

Yes that is what I got

Answer 26

Ok, now it's correct. We have a2/a1 = 4/5.

Answer 27

Now we need to do a3/a2:

\( \dfrac{3}{2} \div 2\) is what we need to do right?

Answer 28

Yes

Answer 29

\(\dfrac{3}{2} \div 2\)

\(= \dfrac{3}{2} \times \dfrac{1}{2} \)

\( = \dfrac{3}{4}\)

Now we have a3/a2 = 3/4

Answer 30

Now we know that \( \dfrac{a2}{a1} \ne \dfrac{a3}{a2} \)

This means it's not a geometric sequence since there is no common ratio.

Answer 31

Okay. I understand this. Thank you so much for the help! To clarify, when there is no common ration it is not geometric?

Answer 32

Correct.
Now we can test it to see if it's an arithmetic sequence.

Answer 33

Oh thank you but you do not have to do that. I only needed to see if it was geometric or not.

Answer 34

Ok, but just in case, here's the info for future reference.

For an arithmetic sequence, you need a common difference.
Instead of dividing each term by the previous term looking for a common ratio, you subtract each term from the next term looking for a common difference. If there is a common difference then it is an arithmetic sequence.

Answer 35

In fact, in this case it is an arithmetic sequence.

Answer 36

You're welcome.

Answer 37

Thank you for the information for future reference, but I actually know how to figure that out already. The extra help is appreciated though. :)

Answer 38

You're welcome.

Answer 39

Good night and thanks again.

Answer 40

Good night, and you're welcome again.