- #1

brotherbobby

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- 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)

**Let me copy and paste the problem as it appears in the text to the right.
Problem statement : **

**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?

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