# Continued proportion problem

#### kuheli

##### New member
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 !!!

#### Sudharaka

##### Well-known member
MHB Math Helper
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 !!!
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.