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mean proportion problem

kuheli

New member
Oct 27, 2013
14
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
Feb 24, 2012
13,775
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
Oct 27, 2013
14
i am unable to figure out in which direction and how to proceed with this
 

soroban

Well-known member
Feb 2, 2012
409
Hello, kuheli!

[tex]\text{If }b\text{ is the mean proportion between }a\text{ and }c,[/tex]

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

We have: .[tex]\frac{a}{b} \,=\,\frac{b}{c} \quad\Rightarrow\quad b^2 \,=\,ac[/tex]

The right side is:

. . [tex](ab+bc + ac)^3 \;=\; (ab + bc + b^2)^3[/tex]

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

. . [tex]=\;b\!\cdot\!b^2(a+b+c)^3 \;=\;b\!\cdot\!ac(a+b+c)^3[/tex]

. . [tex]=\;abc(a+b+c)^3[/tex]
 

kuheli

New member
Oct 27, 2013
14
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
Feb 2, 2012
409
Hello again, kuheli!

[tex]\text{If }a,b,c,d \text{ are in continued proportion,}[/tex]

[tex]\text{prove that: }\;\frac{a}{d} \:=\:\frac{a^3 +b^3 +c^3}{b^3 + c^3 +d^3}[/tex]

I assume that continued proportion means geometric sequence.

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

The left side is: .[tex]\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}[/tex]

The right side is: .[tex]\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} [/tex]

. . . . [tex]=\;\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)}[/tex]

. . . . [tex]=\;\frac{1}{r^3}[/tex]


Q.E.D.
 

kuheli

New member
Oct 27, 2013
14
Hello again, kuheli!


I assume that continued proportion means geometric sequence.

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

The left side is: .[tex]\frac{a}{d} \:=\:\frac{a}{ar^3} \:=\:\frac{1}{r^3}[/tex]

The right side is: .[tex]\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} [/tex]

. . . . [tex]=\;\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)}[/tex]

. . . . [tex]=\;\frac{1}{r^3}[/tex]


Q.E.D.
thanks a lot ...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775