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Ernesto's question at Yahoo! Answers regarding harmonic and arithmetic progressions
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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.
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.