# mean proportion problem

#### kuheli

##### New member
if b is the mean proportion between a and c ,show that

abc(a+b+c)^3 = (ab+bc+ca)^3

#### MarkFL

##### Administrator
Staff member
I have edited your thread title to indicate what kind of question is being asked. A title such as "please help" does not give any information regarding the question.

Also, can you show us what you have tried so our helpers know where you are stuck and can best help?

#### kuheli

##### New member
i am unable to figure out in which direction and how to proceed with this

#### soroban

##### Well-known member
Hello, kuheli!

$$\text{If }b\text{ is the mean proportion between }a\text{ and }c,$$

$$\text{show that: }\:abc(a+b+c)^3 \:=\: (ab+bc+ca)^3$$

We have: .$$\frac{a}{b} \,=\,\frac{b}{c} \quad\Rightarrow\quad b^2 \,=\,ac$$

The right side is:

. . $$(ab+bc + ac)^3 \;=\; (ab + bc + b^2)^3$$

. . $$=\;\big[b(a+c +b)\big]^3 \;=\; b^3(a+b+c)^3$$

. . $$=\;b\!\cdot\!b^2(a+b+c)^3 \;=\;b\!\cdot\!ac(a+b+c)^3$$

. . $$=\;abc(a+b+c)^3$$

#### kuheli

##### New member
hi ... thanks a lot .
there is another problem with which i am stuck .i tried but its getting more complicated and long without any positive sign.the numerical is:

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

a/d=(a^3 +b^3 +c^3)/(b^3 + c^3 +d^3)

#### soroban

##### Well-known member
Hello again, kuheli!

$$\text{If }a,b,c,d \text{ are in continued proportion,}$$

$$\text{prove that: }\;\frac{a}{d} \:=\:\frac{a^3 +b^3 +c^3}{b^3 + c^3 +d^3}$$

I assume that continued proportion means geometric sequence.

So we have: .$$\begin{Bmatrix}a &=& a \\ b &=& ar \\ c &=& ar^2 \\ d &=& ar^3 \end{Bmatrix}$$ .where $$r$$ is the common ratio.

The left side is: .$$\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}$$

The right side is: .$$\frac{a^3+b^3 + c^3}{b^3+c^3+d^3} \:=\:\frac{a^3 + (ar)^3 + (ar^2)^3}{(ar)^3 + (ar^2)^3 + (ar^3)^3}$$

. . . . $$=\;\frac{a^3 + a^3r^3 + a^3r^6}{a^3r^3+a^3r^6+a^3r^9} \;=\; \frac{a^3(1+r^3+r^6)}{a^3r^3(1+r^3+r^6)}$$

. . . . $$=\;\frac{1}{r^3}$$

Q.E.D.

#### kuheli

##### New member
Hello again, kuheli!

I assume that continued proportion means geometric sequence.

So we have: .$$\begin{Bmatrix}a &=& a \\ b &=& ar \\ c &=& ar^2 \\ d &=& ar^3 \end{Bmatrix}$$ .where $$r$$ is the common ratio.

The left side is: .$$\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}$$

The right side is: .$$\frac{a^3+b^3 + c^3}{b^3+c^3+d^3} \:=\:\frac{a^3 + (ar)^3 + (ar^2)^3}{(ar)^3 + (ar^2)^3 + (ar^3)^3}$$

. . . . $$=\;\frac{a^3 + a^3r^3 + a^3r^6}{a^3r^3+a^3r^6+a^3r^9} \;=\; \frac{a^3(1+r^3+r^6)}{a^3r^3(1+r^3+r^6)}$$

. . . . $$=\;\frac{1}{r^3}$$

Q.E.D.
thanks a lot ...

Staff member