+118609 Hi AartarikRoy :)
a,b,c are in A.P , a^2,b^2 and c^2 are in H.P. Show that a=b=c.
I had to look up what a Harmonic Progression was - I've never heard the term before.
A Harmonic Progression is a sequence of quantities whose reciprocals are in arithmetical progression (e.g. 1, 1/3, 1/5, 1/7, etc.).
\(b-a=c-b\qquad(1)\\ 2b=c+a\qquad(1b)\\ 4b^2=c^2+a^2+2ac \qquad(1c)\\ and\\ \frac{1}{b^2}-\frac{1}{a^2}=\frac{1}{c^2}-\frac{1}{b^2} \qquad \qquad (2)\\ \frac{2}{b^2}=\frac{1}{c^2}+\frac{1}{a^2} \qquad \qquad (2b)\\ 2a^2c^2=a^2b^2+b^2c^2 \qquad \qquad (2c)\\ 2a^2c^2=b^2(a^2+c^2) \qquad \qquad (2d)\\ 8a^2c^2=(c^2+a^2+2ac)(a^2+c^2) \qquad \qquad (3)\\ \)
Wolfram|Alpha says the only answer to this is that a=c and by substitution b=a=c
http://www.wolframalpha.com/input/?i=8a%5E2c%5E2%3D(c%2Ba)%5E2(c%5E2%2Ba%5E2)
Of hand I do not yet see how to prove this manually. ://
