Just curious about why geometric progression is called so. Is it related to geometry?

Martin Sleziak
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  • See, https://uk.answers.yahoo.com/question/index?qid=20110329104007AAwq7GY and http://www.quora.com/Why-is-geometric-progression-called-so – lab bhattacharjee May 14 '15 at 13:01
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    See also the Geometric Mean entry from the ["Earliest Known Uses of Some of the Words of Mathematics" site](http://jeff560.tripod.com/g.html). Unfortunately, this doesn't explain the "why", but it provides some leads for further investigation. – Blue May 14 '15 at 16:06
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    @Blue: kudos for a useful reference to the actual history. – Rob Arthan May 14 '15 at 19:11
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    Older questions that missed out on becoming popular: [Arithmetic and geometric sequences: where does their name come from?](http://math.stackexchange.com/q/820680) and [Origin of the words arithmetic and geometric progression](http://math.stackexchange.com/q/967456) – epimorphic Jun 08 '15 at 15:37
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    That linked question links back here to provide more details, so I don't think this one should be closed. – Kevin Feb 16 '21 at 22:23

6 Answers6


Here's an extremely nice picture illustrating the geometric series and making it geometrically clear how it converges when $x<1$.

Dobbs 1918

$s-1=xs$ follows from $PN/ON=BA/OA$ which follows from the fact that $\triangle OAB$ is similar to $\triangle ONP$.

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Gregory Grant
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    Great added bonus for the (unfamiliar to me) proof for the sum :) – nbubis May 15 '15 at 04:11
  • @nbubis Right, I only knew how to make the geometric series intuitive and visual for $x=1/2$, but this allows one to do it for all $x\in(0,1)$. – Gregory Grant May 15 '15 at 17:47
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    What book is this? – Alec Teal May 16 '15 at 18:13
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    @AlecTeal It's this one: http://www.amazon.com/Plane-Trigonometry-H-S-Carslaw/dp/B00BGGHADM/ref=tmm_hrd_title_0 – Gregory Grant Jun 07 '15 at 12:51
  • @AlecTeal I realized you probably meant what book is the one that is referencing Carslaw. I'm not sure, I found it here: http://imgkid.com/geometric-sequences-formulas.shtml And I can't seem to find any other reference to it. It's pretty obscure I guess. – Gregory Grant Jun 07 '15 at 14:05
  • @Gregory Grant: Really beautiful. And incidentally also "makes" analysis, not just geometry. – Piquito Jun 30 '15 at 12:38
  • wow ...you da man...finally I understand – Sedumjoy Jan 09 '18 at 21:28
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    @AlecTeal The image is not from a book; it comes from page 245 of this paper: W.J. Dobbs, "The Introduction to Infinite Series", _The Mathematical Gazette_ 9, no. 135 (May 1918), pp. 242--246, DOI: [10.2307/3605310](http://dx.doi.org/10.2307/3605310). – Senex Feb 14 '19 at 14:18
  • @AlecTeal Thanks for letting us know. – Gregory Grant Feb 15 '19 at 17:43

Here is a geometric figure illustrating the geometric progression $1,r,r^2,r^3,r^4,r^5,\ldots$:

enter image description here

David K
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    Now that's a good picture! We don't have to worry about whether people were thinking about $8$-dimensional cubes, however long ago the terminology came about. – pjs36 May 14 '15 at 14:26
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    So I guess we pick $r$ and draw the edge you have labelled as $r$. Then the hypotenuse of the resulting triangle is $r^2$? Can you please post a proof of that? Does this depend on the length of the other two edges of the first right triangle (the leftmost one with one leg labelled $1$). You haven't labelled them and they aren't uniquely determined by one leg being length one. Thank you. – Gregory Grant May 14 '15 at 14:37
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    @GregoryGrant The triangles are all similar, that uniquely determines everything. The second triangle from the bottom is drawn first and then the first triangle is drawn similar to the second and so on... – Asvin May 14 '15 at 14:47
  • I see it now, thanks – Gregory Grant May 14 '15 at 14:55
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    @GregoryGrant Its pretty easy to proof, you can prove it using the properties of similar triangles. Just do an internet search and you will definitely find a very neat proof. ;) – dark32 May 14 '15 at 15:24
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    Yes I proved it, the conceptual leap was to realize the second triangle from the bottom is drawn first. Then it's easy to show that all right triangles in this figure are similar which follows from Euclid's most basic propositions on angles and parallel lines. – Gregory Grant May 14 '15 at 15:46
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    Is there any useful information that can be read off this figure? What would it look like for the interesting case when $r < 1$? – Rob Arthan May 14 '15 at 19:16
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    @RobArthan If $r<1$, you draw the figure approaching the vertex rather than going away from it. – Mario Carneiro May 15 '15 at 09:13
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    Beautiful pictorial illustration of the geometric progression when $r\gt1$. However this is not an answer to the question about the provenance of the term "geometric". – Piquito Jun 30 '15 at 03:26
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    @Ataulfo I've since become aware of this paragraph from the Oxford English Dictionary: "_Arithmetical progression, proportion, ratio,_ etc. ... relate to differences instead of quotients. The term _geometrical_ points to the fact that problems involving multiplication were originally dealt with by geometry and not by arithmetic." That's the best answer I have for the first part of the question. – David K Jun 30 '15 at 04:46
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    @David K: That´s right. Looking for information about this question I find that hypergeometric series (the prefix “hyper” is well known) are called so because they generalize the geometric series; Pfaff was the mathematician (teacher and friend of Gauss and later Möbius teacher) who introduced this term. – Piquito Jun 30 '15 at 12:29

Let $ABC$ is a right triangle with right angle $\angle ABC$, then if we draw the height $BH$, we have $$ |BH|^2=|AH||CH| $$ The geometric mean comes from here...

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The arithmetic and geometric adjectives come from the Pythagoreans before the Christian Era. Apparently, the expression “geometric progression” comes from the “geometric mean” (Euclidean notion) of segments of length $a$ and $b$: it is the length of the side $c$ of a square whose area is equal to the area of the rectangle of sides $a$ and $b$. The construction of the geometric mean with ruler and compass is well known for high school students; it involves "multiplication" and not "addition".

enter image description here

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    This picture perfectly explains why the geometric mean is smaller then the arithmetic mean if $a\neq b$. and equal if a=b. Beautiful. By Thales' theorem its the same as the comment of the right triangle – Libertas Aug 08 '20 at 06:53
  • Thank you for your comment, dear friend. – Piquito Aug 09 '20 at 11:08

My guess would be that geometric sequences arose as generalization of sequence $a,a^2,a^3,...$. Why is this geometric? Well, $a$ is the length (= 1-dimensional "volume") of line segment (= 1-dimensional hypercube) of "side" length $a$, $a^2$ is the area (= 2-dimensional "volume") of a square (= 2-dimensional hypercube) of side length $a$, $a^3$ is the volume (= 3-dimensional "volume") of a cube (= 3-dimensional hypercube) of side length $a$ and so on.

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Because geometric progressions are based on multiplication, and the most important geometric notion, namely, volume, arises from multiplication (length times width times height). The term “multiplicative” is not used because it already has a special meaning in Number Theory.

Mike Jones
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