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Formulas for Arithmetical and Geometrical Progression 
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#1
Apr2514, 10:45 AM

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Memorization, a contemptuous, futile, and insignificant method of learning math is, I think, not the solution to my problem. Well for the arithmetical progression, I have the formula:
Let: l= last term a= first term d= common difference l= a+[d(n1)] And for the geometrical progression; Let: l= last term a= first term r= common ratio l= a(r^n1) and also, for the getting the sum; Let: l= last term r= common ratio a= first term S= lra/ r1 Now given these formulas, I have no idea how it is derived. I tried to formulate it on my own but I don't know how these equation was created. Please, give me tips to discern how a formula is made. 


#2
Apr2514, 10:52 AM

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#3
Apr2514, 10:58 AM

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You can make a very similar argument for the geometric progression: the first term is ##a##, the second term is ##ar##, the third term is ##ar^2##, and in general, the ##n##'th term is ##ar^{n1}##. The sum of the geometric progression is more interesting. Here, if you want to sum the first ##n## terms, you need to evaluate ##S = a + ar + ar^2 + \ldots + ar^{n1}##. All of the terms have a common factor ##a##, so we can factor it out to get ##S = a(1 + r + r^2 + \ldots + r^{n1})##. So the key is to find $$T = 1 + r + r^2 + \ldots + r^{n1}$$ Note that we can multiply the entire equation by ##r## to obtain $$rT = r + r^2 + \ldots + r^{n1} + r^n$$ Now (great trick) subtract the second equation from the first, to obtain $$T  rT = 1  r^n$$ since all the other terms cancel. Now divide both sides by ##1r## to get $$T = \frac{1  r^n}{1r}$$ Thus $$S = aT = a \frac{1  r^n}{1r}$$ is the answer you want. Note that it is valid as long as ##r \neq 1##. Can you find another formula for the ##r=1## case? 


#4
Apr2514, 11:04 AM

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Formulas for Arithmetical and Geometrical Progression
I never saw it that way, pretty interesting stuff. Thanks! What is the importance of series? or infinite series by the way? Sorry if I sound so ignorant since I really am. I heard its really important in higher mathematics. 


#5
Apr2514, 11:07 AM

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When I first read your post I didn't notice you were also asking about the SUM of the geometric progression, which is more interesting. So I added some info about that to my previous response.



#6
Apr2514, 11:10 AM

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#7
Apr2514, 11:17 AM

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#8
Apr2514, 11:18 AM

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#9
Apr2514, 11:27 AM

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By the way, the geometric series has (surprise!) a nice geometric interpretation, for example when ##r=1/2##. The sum ##S = 1/2 + 1/4 + 1/8 + 1/16 + \ldots## adds up to ##1## in the limit, because it is formed by adding half of the unit square (1/2), then half of the remaining area (1/4), then half of the remaining area (1/8), etc. See the pictures here:
https://sites.google.com/site/butwhy...eriesvisually 


#10
Apr2514, 11:39 AM

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#11
Apr2514, 11:48 AM

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#12
Apr2514, 11:50 AM

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#13
Apr2514, 12:13 PM

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Once you get the general idea,it will be as easy as pie. 


#14
Apr2514, 12:19 PM

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Oh, of course, I know that visualization, but in some ways, for example, dividing a fraction by a fraction, I can't visualize it through a circle or a square but through a number line only. But I think I got the idea of adding a unit fraction by a unit fraction visually.



#15
Apr2514, 12:51 PM

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For example, instead of, say (1/2)/(1/3), this is the same as (1/2) * (3/1), or 3/2. 


#16
Apr2514, 01:26 PM

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I didn't mean that. I meant to say, I can't visualize dividing a fraction by a fraction with the use of a figure, or shape, like circles or squares, but I can do the operation, abstractly, using numbers. Although I can picture it using the number line. And also, I can't visualize multiplying a fraction by a fraction using a model, using a figure, a shape, such as circle and square.



#17
Apr2514, 01:43 PM

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$$\frac{1/2}{1/8} = 4$$ In my opinion, trying to visualize it using squares or circles introduces unnecessary clutter. 


#18
Apr2514, 01:48 PM

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Never mind, guys, I completely see now visually how multiplication of fractions works using shapes.



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