Alg9.4.22

If $\log(a+c)+\log(a-2b+c)=2\log(a-c)\,$ then show that a,b,c are in Harmonic Progression.

Simplifying the equation,we get

$\log((a+c)\cdot(a-2b+c))=\log(a-c)^{2}\,$

Taking out the logs on both sides,

$(a+c)\cdot(a-2b+c)=(a-c)^{2}\,$

$a^2-2ab+ac+ac-2bc+c^2=a^2+c^2-2ac\,$

Simplifying further,

$4ac=2bc+2ab\,$

$2ac=b(c+a)\,$

$b=\frac{2ac}{c+a}\,$

a,b,c are in harmonic progression if $b=\frac{2ac}{c+a}\,$

Hence a,b,c are in harmonic progression.