If ln(a+c), ln(c-a), ln(a-2b+c) are in A.P.,t | vedclass

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(D) Given that $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$Therefore,$(a+c), (c-a), (a-2b+c)$ are in $G.P.$This implies $(c-a)^2 = (a+c)(a-2b+c)$.E

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

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(D) Given that $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$Therefore,$(a+c), (c-a), (a-2b+c)$ are in $G.P.$This implies $(c-a)^2 = (a+c)(a-2b+c)$.Expanding the terms: $c^2 - 2ac + a^2 = a^2 - 2ab + ac + ac - 2bc + c^2$.Simplifying: $-2ac = -2ab + 2ac - 2bc$.Rearranging: $2ab + 2bc = 4ac$.Dividing by $2b(a+c)$: $b = \frac{2ac}{a+c}$.This is the condition for $a, b, c$ to be in $H.P.$

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

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