How to Prove a Mean Proportion Problem in Algebra?

[kuheli]

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

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

 

[MarkFL]

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]

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

 

[soroban]

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]

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]

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]

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 ...

 

[MarkFL]

thanks a lot ...

I have moved your next question to its own thread here:

http://mathhelpboards.com/pre-algebra-algebra-2/continued-proportion-problem-7531.html

It is better to create create new threads for new questions. You are more likely to get prompt help this way and our threads do not become long strings of successive problems which are more difficult to follow.