MHB Solving Continued Proportion Problem with Componendo Dividendo

[kuheli]

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 into (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 !

 

[Sudharaka]

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 into (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 !

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. :)