Ernesto's Question: Proving Harmonic & Arithmetic Progressions

[MarkFL]

Here is the question:

Prove that if b+c, a+c, a+b are in harmonic progression, then a², b² y c²...?


Prove that if b+c, a+c, a+b are in harmonic progression, then a², b² y c² are in arithmetic progression .

Explain.

Thanks.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Ernesto,

If $b+c,\,a+c,\,a+b$ are in harmonic progression, then their reciprocals are in arithmetic progression, and we may state:

(i) \(\displaystyle \frac{1}{a+c}=\frac{1}{b+c}+d\)

Multiply through by \(\displaystyle (a+c)(b+c)\) to obtain:

\(\displaystyle b+c=a+c+d(a+c)(b+c)\)

Subtract $c$ from both sides and expand the product on the far right:

\(\displaystyle b=a+d(ab+ac+bc+c^2)\)

Distribute on the right:

\(\displaystyle b=a+abd+acd+bcd+c^2d\)

Subtract through by \(\displaystyle a+abd+acd+bcd\):

\(\displaystyle b-a-abd-acd-bcd=c^2d\)

Divide through by $d$ and arrange as:

\(\displaystyle c^2=\frac{b-a-abd-acd-bcd}{d}\)

(ii) \(\displaystyle \frac{1}{a+b}=\frac{1}{b+c}+2d\)

Multiply through by \(\displaystyle (a+b)(b+c)\) to obtain:

\(\displaystyle b+c=a+b+2d(a+b)(b+c)\)

Subtract $b$ from both sides and expand the product on the far right:

\(\displaystyle c=a+2d(ab+ac+b^2+bc)\)

Distribute on the right:

\(\displaystyle c=a+2abd+2acd+2b^2d+2bcd\)

Subtract through by \(\displaystyle a+2abd+2acd+2bcd\):

\(\displaystyle c-a-2abd-2acd-2bcd=2b^2d\)

Divide through by $2d$ and arrange as:

\(\displaystyle b^2=\frac{c-a-2abd-2acd-2bcd}{2d}\)

(iii) \(\displaystyle \frac{1}{a+b}=\frac{1}{a+c}+d\)

Multiply through by \(\displaystyle (a+b)(a+c)\) to obtain:

\(\displaystyle a+c=a+b+d(a+b)(a+c)\)

Subtract $a$ from both sides and expand the product on the far right:

\(\displaystyle c=b+d(a^2+ac+ab+bc)\)

Distribute on the right:

\(\displaystyle c=b+a^2d+acd+abd+bcd\)

Subtract through by \(\displaystyle b+acd+abd+bcd\):

\(\displaystyle c-b-acd-abd-bcd=a^2d\)

Divide through by $d$ and arrange as:

\(\displaystyle a^2=\frac{c-b-acd-abd-bcd}{d}\)

Now in order for $a^2,\,b^2,\,c^2$ to be in arithmetic progression, we require:

\(\displaystyle b^2-a^2=c^2-b^2\)

Add through by \(\displaystyle a^2+b^2\):

\(\displaystyle 2b^2=a^2+c^2\)

Substitute for $a^2,\,b^2,\,c^2$ the expressions we found above:

\(\displaystyle 2\left(\frac{c-a-2abd-2acd-2bcd}{2d} \right)=\frac{c-b-acd-abd-bcd}{d}+\frac{b-a-abd-acd-bcd}{d}\)

Distributing the 2 on the left, and the multiplying through by $d$, we find:

\(\displaystyle c-a-2abd-2acd-2bcd=c-b-acd-abd-bcd+b-a-abd-acd-bcd\)

Add through by \(\displaystyle 2abd+2acd+2bcd\):

\(\displaystyle c-a=c-b+b-a\)

Collect like terms:

\(\displaystyle c-a=c-a\)

Subtract through by \(\displaystyle c-a\):

\(\displaystyle 0=0\)

This is an identity, which proves that given $b+c,\,a+c,\,a+b$ are in harmonic progression, then $a^2,\,b^2,\,c^2$ must be in arithmetic progression.