If ln(a + c),ln(c - a),and ln(a - 2b + c) are | vedclass

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(D) Given that $\ln(a + c)$,$\ln(c - a)$,and $\ln(a - 2b + c)$ are in $A.P.$Therefore,$2\ln(c - a) = \ln(a + c) + \ln(a - 2b + c)$Using the property $

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$a, b, c$ are in $H.P.$

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(D) Given that $\ln(a + c)$,$\ln(c - a)$,and $\ln(a - 2b + c)$ are in $A.P.$Therefore,$2\ln(c - a) = \ln(a + c) + \ln(a - 2b + c)$Using the property $\ln(x) + \ln(y) = \ln(xy)$,we get:$\ln((c - a)^2) = \ln((a + c)(a - 2b + c))$Removing the logarithms:$(c - a)^2 = (a + c)(a - 2b + c)$$c^2 + a^2 - 2ac = a^2 - 2ab + ac + ac - 2bc + c^2$$c^2 + a^2 - 2ac = a^2 + c^2 + 2ac - 2ab - 2bc$$-2ac = 2ac - 2b(a + c)$$2b(a + c) = 4ac$$b = \frac{2ac}{a + c}$This is the condition for $a, b, c$ to be in $H.P.$

If $\ln(a + c)$,$\ln(c - a)$,and $\ln(a - 2b + c)$ are in $A.P.$,then

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