Question

Which of the following are geometric sequences?

3,6,9,12,15,18
5,10,20,40,80,160
10,5,2.5,1.25,0.625,0.3125
1,3,9,27,81

Answer 1

Calculate successive ratios. I'll do the first one.

3,6,9,12,15,18

6/3 = 2
9/6 = 1.5 -- Whoops! We can stop there. As soon as we get a second answer, it's NOT a geometric sequence.

Okay, you show me the other three.

Answer 2

Oh, so I'm just seeing if the ratios are the same between each number?
So would the second and fourth sequences be geometric than?

Answer 3

You tell me. What is the common ratio?

Give #3 some more thought.

Answer 4

Ah! Number three is geometric too. For the second set the common ratio is 2? For the third, .5? For the fourth, 3?

Answer 5

I think you're done. Good work.

Answer 6

Could you help with a different problem too? I would like to know how to do these since I have a lot of them to do..

What is the sum of the first five terms of a geometric series with a1 = 20 and r = 1/4?

Answer 7

Seriously? There are only 5 terms. Why not just add them up?

What ARE the first 5 terms?

Answer 8

I guess I'm confused on what to do with the r=1/4 part....

Answer 9

Take a look at the original sequence #3. What did that do with the 1/2?

Answer 10

So, I'd multiply 20 by .25, than mulitply 5 by .25 and so on?

Answer 11

For the sum, I got about 6.65?

Answer 12

That's the idea. Is that right?

Answer this, what is the sum of ALL the terms of the sequence if we do NOT stop with the 5th?

Answer 13

Would the sum be 20 since that was the number started with? I'm sure that's not the right answer though

Answer 14

20 + 5 = 25
20 + 5 + 5/4 = 105/4 = 26.25
20 + 5 + 5/4 + 5/16 = 425/16 = 26.5625
20 + 5 + 5/4 + 5/16 + 5/64 = 1705/64 = 26.640625

That is the answer, but it there an easier way to get too it? That was pretty annoying!

You should have two important formulas.

The sum of infinitely many terms of a geometric series. If it has a sum, that is.

\(Sum_{\infty} = \dfrac{a_{0}}{1-r}\)

Does that look familiar?

Answer 15

Kind of, I missed class the day we were taught all this stuff, so I apologize for not being that bright. Since it asks me to turn it into an improper fraction, would I just use the 1705/64 part?

Answer 16

Well, it's important to see it, so I will show you, quickly.

You can use that same formula to find any other sum of any infinite number of later sequential terms.

\(\dfrac{a_{5}}{1-r}\) is the sum of all terms of the sequence, starting at the 6th. (Remember, we started counting at 0, so the index of 5 is the 6th value. It is a little confusing.)

Thus, if we want to know the sum of the first 5 terms, we can us this idea, almost magic!

\(\dfrac{a_{0}}{1-r} - \dfrac{a_{5}}{1-r}\) This is literally the sum of all terms and then subtracting the sum of all terms after the 5th. A little algebra gives:

\(\dfrac{a_{0}-a_{r}}{1-r} = \dfrac{a_{0} - a_{0}\cdot r^{5}}{1-r} = a_{0}\cdot\dfrac{1-r^{5}}{1-r}\)

Thus, and here is the better way, with \(a_{0} = 20\;and\;r = \dfrac{1}{4}\),

\(20\cdot\dfrac{1-\left(\dfrac{1}{4}\right)^{5}}{1-\dfrac{1}{4}} = 20\cdot\dfrac{1 - \dfrac{1}{1024}}{\dfrac{3}{4}} = 20\cdot\dfrac{\dfrac{1023}{1024}}{\dfrac{3}{4}} = 20\cdot\dfrac{1023}{1024}\cdot\dfrac{4}{3} =\)


\(5\cdot\dfrac{1023}{3}\cdot\dfrac{16}{1024} = 5\cdot 341\cdot \dfrac{1}{64} = \dfrac{1705}{64}\)

That was mostly arithmetic.

Answer 17

Yes that definitely was. I will screen shot that to remember. Thank you so much!

Answer 18

It's a beautiful thing.