A harmonic progression is a sequence of real numbers formed by taking the reciprocals of an arithmetic progression. Equivalently, it is a sequence of real numbers such that any term in the sequence is the harmonic mean of its two neighbors.
$a,b,c$ are in HP
$\implies \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in AP
$\implies \dfrac{1}{b} – \dfrac{1}{a} = \dfrac{1}{c} – \dfrac{1}{b}$
$\implies \dfrac{1}{b} + \dfrac{1}{b} = \dfrac{1}{c} + \dfrac{1}{a}$
$\implies \dfrac{2}{b} = \dfrac{a+c}{ac}$
$\implies \dfrac{b}{2} = \dfrac{ac}{a+c}$
$\implies b = \dfrac{2ac}{a+c}$
$\implies 2b(a+c) = 4ac $
$\log(a+c)+\log(a−2b+c) = \log[(a+c)(a+c-2b)]$
$\implies\log(a+c)+\log(a−2b+c) = \log[(a+c)^{2}-2b(a+c)]$
$\implies\log(a+c)+\log(a−2b+c) = \log[(a+c)^{2}-4ac]$
$\implies\log(a+c)+\log(a−2b+c) = \log(a - c)^{2}$
$\implies\log(a+c)+\log(a−2b+c) = 2\log(c - a)\:\:\:\:(\because c>a)$
Reference:https://brilliant.org/wiki/harmonic-progression/
So, the correct answer is $(C).$