MHB Solving Ratio and Proportion Problem: (a+b+c)^2/(a^2+b^2+c^2)

[kuheli]

hi ,

i am stuck with a problem . the problem is

if a , b , c are in continued proportion ,then prove that

(a+b+c)^2/(a^2 +b^2 +c^2) =(a+b+c)/(a-b+c)

i have tried solving the problem in different way like breaking the formula of (a+b+c)^2 =a^2 +b^2 +c^2 + 2ab +2bc+2ca , then used componendo divedendo but ultimately no success. please help to solve the problem ...

 

[Evgeny.Makarov]

Re: ratio and proportion

When you look at the https://driven2services.com/staging/mh/index.php?posts/34268/ to a similar problem, what would you say is the first step?

 

[kuheli]

Re: ratio and proportion

what ? i am not getting you ...

 

[Evgeny.Makarov]

Well, maybe you can say what it means for $a$, $b$, $c$ to be in continued proportion.

 

[kuheli]

a,b,c in continued proportion means a/b=b/c
i.e b^2=ac

 

[Evgeny.Makarov]

a,b,c in continued proportion means a/b=b/c
i.e b^2=ac

Yes. Another way to look at this is to denote $b/a=r$, i.e., $a/b=1/r$. Then $b=ar$, and $a/b=b/c$ gives $1/r=b/c$, i.e., $c=br$. That is, \[
b=ar\tag{1}
\]
and \[
c=br=a^2r\tag{2}.
\]
Now replace $c$ and $b$ in the equation you need to prove using (1) and (2), so that the only variables left are $a$ and $r$.

 

[kuheli]

i tried that way.no result found

 

[Evgeny.Makarov]

Let's see. You need to prove
\[
\frac{(a+ar+ar^2)^2}{a^2+(ar)^2+(ar^2)^2} = \frac{a+ar+ar^2}{a-ar+ar^2}
\]
First you can cancel $a+ar+ar^2$, which gives
\[
\frac{a+ar+ar^2}{a^2+(ar)^2+(ar^2)^2} = \frac{1}{a-ar+ar^2}
\]
Second, factor out all $a$'s and cancel them.
\[
\frac{a(1+r+r^2)}{a^2(1+r^2+r^4)} = \frac{1}{a(1-r+r^2)}
\]
i.e.,
\[
\frac{1+r+r^2}{1+r^2+r^4} = \frac{1}{1-r+r^2}
\]
Now multiply across (i.e., multiply both sides by both denominators) and represent $(1+r+r^2)(1-r+r^2)$ as $((1+r^2)+r)((1+r^2)-r)$ to use the formula $(x+y)(x-y)=x^2-y^2$.

Can you finish?