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

[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?

 

[martinbn]

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.

 

[BvU]

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

[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.

##\ ##

 

[PeroK]

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.

 

[WWGD]

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.

 

[PeroK]

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!

 

[brotherbobby]

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.

 

[brotherbobby]

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}\\
\boxed{a^2(b+c),b^2(c+a),c^2(a+b)}&\rightarrow \text{in AP} \quad\color{green}{\huge\checkmark}\\
\end{split}
\end{equation*}