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If three positive real numbers $a, b$ and $c(c>a)$ are in Harmonic Progression, then $\log\left ( a+c \right )+\log\left ( a-2b+c \right )$ is equal to:

  1. $2\:\log\left ( c-b \right )$
  2. $2\:\log\left ( a-c\right )$
  3. $2\:\log\left ( c-a\right )$
  4. $\log\:a+\log\:b+\log\:c$
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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 $(C).$

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