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

It has been given that the terms $$a, b,$$ and $$c$$ are in harmonic progression.
Therefore, $$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$$
$$\frac{2}{b}$$ = $$\frac{1}{a}$$ + $$\frac{1}{c}$$
$$\frac{2}{b} = \frac{a+c}{ac}$$
$$b = \frac{2ac}{(a+c)}$$--------------(1)
The given expression is log $$(a+c)$$ + log $$(a-2b+c)$$.
log $$(a+c) + log (a - 2b + c)$$ = log $$((a+c)(a-2b + c))$$
Substituting (1), we get,
log $$(a+c)$$ + log $$(a - 2b + c)$$ = log$$((a+c)(a - \frac{4ac}{(a+c)} +c))$$
 = log ($$a^2 + ac - 4ac + c^2 + ac$$)
 = log $$(a^2 + c^2 - 2ac)$$
 = log $$(c-a)^2$$ [Since c is greater than a]
= 2 log $$(c-a)$$
Therefore, option C is the right answer.