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[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]
[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]
thanks a lot ...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.
I have moved your next question to its own thread here:thanks a lot ...