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=times new roman]*

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*}