Turning a series into a geometric series by adding a certain value to each of them

[Femme_physics]

1. The problem statement, all variables and given/known data

http://img833.imageshack.us/img833/681/a1a2.jpg (http://imageshack.us/photo/my-images/833/a1a2.jpg/)

Calculate which number you have to add to a1, a2 and a3 in order to get 3 subsequent numbers in a geometric series

3. The attempt at a solution

Getting a2 and a3 was easy.

Plugging in the values I need for n, I get
a2 = 5
a3 = 13

Okay, now by trial and error I figured out that if I add 3 to all the values, I get 3 numbers that the multiplicative value of 2 is their difference. But, how do I get to that fact without a guessing game of trial and error, I don't know. Can anyone help me?

[Dick]

You can take the first terms of the geometric series to be a, a*r and a*r^2. Take the constant you add to be c. So you want to solve 1+c=a, 5+c=a*r and 13+c=a*r^2 for c. That's three equation in three unknowns. Taking quotients of those equations is probably the easiest way.

[Femme_physics]

So you want to solve 1+c=a, 5+c=a*r and 13+c=a*r^2 for c. That's three equation in three unknowns. Taking quotients of those equations is probably the easiest way.

That makes perfect sense. The first term is just the first term, added c to it. The second term is a multiplicative value, so we need to make sure it's multiplied by a certain value. And with the third term, that value is raised to the power of the value the second term is multiplied by.

Hmmm...

Yes.

I can see the logic.

(btw I'm not sure what you mean by "taking quotients", that's too much of a professional mathematical language for me. Though I do know that quotient is the result of a fraction).

Solving these 3 equations though yields c = -0.5
Did I make a mistake along the way?

http://img199.imageshack.us/img199/4218/eqsolving.jpg (http://imageshack.us/photo/my-images/199/eqsolving.jpg/)

[I like Serena]

Oh here you were! :smile:

Your step from

r = \frac {5+c} {1+c}
to

r = 5

is not ok.

This would only be true if c would be zero.

[Femme_physics]

Good point. I'll work on that.

[Femme_physics]

http://img51.imageshack.us/img51/17/fixedp.jpg (http://imageshack.us/photo/my-images/51/fixedp.jpg/)

Nailed it :)

Though I got c=-1 in addition to c=3. But I'll just treat -1 as an extraneous root.

Thanks!

PS Do you like my new way of posting equations? :)

[I like Serena]

Nailed it :)

Yep! :smile:

Though I got c=-1 in addition to c=3. But I'll just treat -1 as an extraneous root.

Thanks!

You should check it. Perhaps there are 2 solutions.
Then you could baffle everybody with your second solution! :wink:
Which solution would you get with c=-1?

PS Do you like my new way of posting equations? :)

Yes! It looks more professional and is easier to read.
Although I kind of liked the personal touch of your handwritten scans with for instance a bug that *bolted*! :smile:
When will I get to see something like that now? :(

[Femme_physics]

You should check it. Perhaps there are 2 solutions.
Then you could baffle everybody with your second solution!
Which solution would you get with c=-1?

Well, a1 would equal 0, so no matter what you multiply it by it's gonna equal zero, so that can't be an answer!

Yes! It looks more professional and is easier to read.
Although I kind of liked the personal touch of your handwritten scans with for instance a bug that *bolted*!
When will I get to see something like that now? :(

Oh, this is only my "work form"! :) That's how you'll know I'm posting from work. Whenever I'll have a scanner I'll go back to handwriting and scanning!

(and oh yea, the bolting spider, that was fun heh :D )

[I like Serena]

Well, a1 would equal 0, so no matter what you multiply it by it's gonna equal zero, so that can't be an answer!

Yep! That's it! :smile:

Note that it's important to check solutions against the original equation.

Oh, this is only my "work form"! :) That's how you'll know I'm posting from work. Whenever I'll have a scanner I'll go back to handwriting and scanning!

What kind of "work form" is that?
Does it have a name?
And how do you make a picture from it?

(and oh yea, the bolting spider, that was fun heh :D )

Yep!!!!! :smile:

[Femme_physics]

What kind of "work form" is that?
Does it have a name?
And how do you make a picture from it?

Microsoft Equation 3.0 in Powerpoint 2007

I just screenshot the whole thing after I finish typing. It's a heckuva lot easier than using LaTeX, because in LaTeX I keep losing myself over all the brackets and code stuff (if I'd look back to try and see what I wrote, I'd have no idea). Microsoft Equation 3.0 is more like "WYSIWYG"

:)

-Or

[I like Serena]

Microsoft Equation 3.0 in Powerpoint 2007

I just screenshot the whole thing after I finish typing. It's a heckuva lot easier than using LaTeX, because in LaTeX I keep losing myself over all the brackets and code stuff (if I'd look back to try and see what I wrote, I'd have no idea). Microsoft Equation 3.0 is more like "WYSIWYG"

You may want to take a look at this one:
http://www.codecogs.com/latex/eqneditor.php

It's an online latex editor, showing your results immediately.
It looks a bit like Microsoft Equation.
And you can copy and paste it easily in PF between [ tex ] tags.

[Femme_physics]

You may want to take a look at this one:
http://www.codecogs.com/latex/eqneditor.php

It's an online latex editor, showing your results immediately.
It looks a bit like Microsoft Equation.
And you can copy and paste it easily in PF between [ tex ] tags.

Terrific! I'll use it, then, since I know it makes the Homework Helper's life and yours easier :)

[Mark44]

Terrific! I'll use it, then, since I know it makes the Homework Helper's life and yours easier :)We will appreciate it. One advantage is that we can insert comments inline, rather than write what you wrote, and then add a correction.