# Prove that terms cyclic in ##a,b,c##, are in ##\text{AP}##

• brotherbobby
brotherbobby
Homework Statement
If ##a,b,c## are in ##\text{AP}##, prove that ##\boxed{a^2(b+c), b^2(c+a), c^2(a+b)}## are also in ##\boxed{\text{AP}}##.
Relevant Equations
If terms ##a,b,c## are in ##\text{AP}##, then ##b-a = c-b = d##, where ##d## is the common difference.
Thus ##b = a+d## and ##c = a+2d##.
(This implies that the required expressions in the box above can be reduced to those having only ##a## and ##d##, eliminating ##b## and ##c## from them)
Problem statement :
Let me copy and paste the problem as it appears in the text to the right.

Attempt : We have the terms of the ##\text{AP}## as ##a, \;b = a+d, \;c = a+2d##
Let the first term of the required expression be ##t_1 = a^2(b+c) = a^2(2a+3d)=2a^3+3a^2d\dots\quad (1)##
Let the second term of the required expression be ##t_2 = b^2(c+a) = (a+d)^2(a+a+2d) = 2(a+d)^2(a+d) = 2a^3+6a^2d+6ad^2+2d^3\dots\quad (2)##.
Let the third term of the expression be ##t_3= c^2(a+b)=(a+2d)^2(2a+d) = 2a^3+9a^2d+12ad^2+4d^3\dots\quad (3)##
(after some algebra).
Subtracting ##(2) - (1)##, we obtain ##t_2-t_1 = 3a^2d+6ad^2+2d^3## and ##(3)-(2)##, ##t_3-t_2 = 3a^2d+6ad^2+2d^3## *
But this is the property of an ##\text{AP}##, whereby ##t_3-t_2=t_2-t_1##.
Hence the required terms ##\boldsymbol{a^2(b+c), b^2(c+a), c^2(a+b)}## are also in ##\mathbf{\text{AP}}##.

* I can't seem to ##\mathrm{\LaTeX}## this expression. Any clues as to why?

Question (doubt) : I completed the proof, but in a way that is not smart. I wrote out the required terms using ##a## and ##d## and did algebra, but did not use the properties of an AP,
For instance, if ##a_1, a_2, \dots, a_n## are ##n## numbers in AP, then we can show that ##a_1\pm k, a_2\pm k, \dots, a_n\pm k## are also in AP. Likewise, we can also show that ##\frac{pa_1}{q}, \frac{pa_2}{q}, \dots, \frac{pa_n}{q}\quad (p,q\ne0)## are also in AP.
To give you a feel of what I mean, I copy and paste below to the right how the author has answered an easier question of the same type :

When I solved this problem before, I again used the method above, viz. express ##c## and ##b## in terms of ##a## and ##d##.

Request : Can someone give a hint of a "smarter" solution to the problem above, whereby I can use the properties of AP and manipulate th variables ##a,b,c## accordingly without resorting to involved algebra?

Last edited:
Another way is to notice that three numbers form an AP if and only if the middle is the average of the other two. So you are given that ##b=\frac{a+c}2## and you need to show that

##b^2(a+c)=\frac{a^2(b+c)+c^2(b+a)}2##

after substitutions this becomes

##\frac{(a+c)^3}4=\frac{a^2(a+3c)+c^2(3a+c)}4##

which can be checked easily. It still needs some algebra though.

PeroK
Re ##\LaTeX ## question: with Reply and toggle BB I see

brotherbobby said:
Code:
[FONT=times new roman]Subtracting ##(2) - (1)##, we obtain ...
##t_3-t_2 = [FONT=times new roman]3a^2d+6ad^2+2d^3## [/FONT]
[COLOR=rgb(255, 0, 0)][FONT=times new roman]*[/FONT][/COLOR][/FONT]
A blunt copy/paste gives
##t_3-t_2 = 3a^2d+6ad^2+2d^3## So the ##\LaTeX ## is ok.

But a change font encloses the second ## and that breaks the math.

##\ ##

SammyS
It must be slightly easier to let ##a = b -d## and ##c = b +d##. Then the new middle term is ##2b^3##. And it is enough to show that
$$(b+d)^2(2b-d) + (b-d)^2(2b+d) = 4b^3$$Which comes out quite easily.

Though, to be overly picky, the author should have stated a,b,c are _ consecutive terms_ in an AP. No reason they can't be the, e.g., 7th, 13th and 59th terms in an AP.

I tried to find a reason why this result holds. First, if ##a, b, c## is an AP with common difference ##d##, then ##a+b, a+c, b+c## is an AP with the same difference ##d##. Now, for any ##x, y## there is a unique number ##z## such that ##(a+b)x, (a+c)y, (b+c)z## is an AP. And, in fact$$z = \frac{2(a+c)y - (a+b)x}{b+c}$$The first thing we might try is setting ##x = c, y = b##, in which case ##z = a + \frac{2d^2}{b + c}##, which is not ##a##.

But, if we set ##x = c^2, y = b^2##, then$$z = \frac{2(a+c)b^2 - (a+b)c^2}{b+c} = a^2$$But, it's still not clear why this simplifies to ##a^2##, other then because that's what the algebra tells you!

WWGD said:
Though, to be overly picky, the author should have stated a,b,c are _ consecutive terms_ in an AP. No reason they can't be the, e.g., 7th, 13th and 59th terms in an AP.
When we say that three numbers ##a,b,c## are in AP, we don't mean they are a part of an AP. If they are the 7th, 13th and 59th terms of an AP, they wouldn't be in AP themselves.

PeroK
Thank you all for your comments. Your methods work and with less algebra than mine. However, after a search on the internet I have found the type of solution I was looking for. It is smart, but one has to know it in advance I suppose, or else not be able to "invent" it. Since I am a bit late with my response, let me start from the beginning.

Problem statement :
Solution :
Please see my attempt in post #1. It was long winded and algebraically involved. I wanted to know if one can use the properties of an AP to solve the problem, keeping the algebra to a minimum.

\begin{equation*}
\begin{split}
a,b,c & \rightarrow \text{in AP} \\
a(ab+bc+ca), b(ab+bc+ca), c(ab+bc+ca) & \rightarrow \text{in AP}\quad\text{(property of AP)}\\
a^2b+abc+a^2c, b^2a+b^2c+abc, abc+c^2b+c^2a & \rightarrow \text{in AP}\\
a^2b+a^2c,b^2a+b^2c, c^2a+c^2b &\rightarrow \text{in AP}\\
\end{split}
\end{equation*}

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## What does it mean for terms to be cyclic in ##a, b, c##?

Terms that are cyclic in ##a, b, c## refer to expressions that remain invariant under the cyclic permutations of ##a, b,## and ##c##. For example, if you have a term ##f(a,b,c)##, then cyclic permutations would include ##f(b,c,a)## and ##f(c,a,b)##.

## What is the definition of an arithmetic progression (AP)?

An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference. For example, in the sequence 2, 5, 8, 11, the common difference is 3.

## How do you prove that cyclic terms in ##a, b, c## are in AP?

To prove that cyclic terms in ##a, b, c## are in AP, you need to show that the difference between consecutive terms is constant. For example, if the terms are ##a, b, c## in cyclic order, you need to demonstrate that ##b - a = c - b = a - c##, which implies a common difference.

## Can you provide an example of cyclic terms in ##a, b, c## that are in AP?

Consider the terms ##a, b, c## such that ##a = 3, b = 7,## and ##c = 11##. These terms are in AP because the differences between consecutive terms are constant: ##b - a = 7 - 3 = 4##, ##c - b = 11 - 7 = 4##, and ##a - c = 3 - 11 = -8##. However, since we are considering cyclic permutations, the common difference should be taken as absolute, hence they form an AP with a common difference of 4.

## Why is it important to understand cyclic terms in AP?

Understanding cyclic terms in AP is important in various fields of mathematics and physics because it helps in identifying symmetrical properties and invariant relationships in equations and systems. This can simplify complex problems and lead to more elegant and general solutions.

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