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Continued proportion problem

kuheli

New member
Oct 27, 2013
14
Re: mean proportion problem

if a,b,c are in continued proportion

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


i break the part -> (a+b+c)^2 in to (a^2 +b^2+c^2+2ab+2bc+2ac) and then use componendo dividendo to the numerator and denominator to the problem,,, but its still not working out .wondering how to solve it . please help !!! :confused:
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: mean proportion problem

if a,b,c are in continued proportion

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


i break the part -> (a+b+c)^2 in to (a^2 +b^2+c^2+2ab+2bc+2ac) and then use componendo dividendo to the numerator and denominator to the problem,,, but its still not working out .wondering how to solve it . please help !!! :confused:
Hi kuheli, :)

Note that,

\[(a-b+c)(a+b+c)=a^2-b^2+c^2+2ac\]

Since \(\frac{a}{b}=\frac{b}{c}\Rightarrow ac=b^2\) we have,

\[(a-b+c)(a+b+c)=a^2+b^2+c^2\]

Hope you can continue. :)