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 $A.P.$By the property of $A.P.$,if $x, y, z$ are in $A.P.$,then $2y = x + z$. Thus:$2 \ln(c-a)

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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 $A.P.$By the property of $A.P.$,if $x, y, z$ are in $A.P.$,then $2y = x + z$. Thus:$2 \ln(c-a) = \ln(a+c) + \ln(a-2b+c)$Using the logarithmic property $\ln(m) + \ln(n) = \ln(mn)$ and $n \ln(m) = \ln(m^n)$:$\ln((c-a)^2) = \ln((a+c)(a-2b+c))$Taking the exponential on both sides:$(c-a)^2 = (a+c)(a-2b+c)$$c^2 - 2ac + a^2 = a^2 - 2ab + ac + ac - 2bc + c^2$$c^2 - 2ac + a^2 = a^2 - 2ab + 2ac - 2bc + c^2$Subtracting $a^2 + c^2$ from both sides:$-2ac = -2ab + 2ac - 2bc$$2ab + 2bc = 4ac$$ab + bc = 2ac$$b(a+c) = 2ac$$b = \frac{2ac}{a+c}$This is the condition for $a, b, c$ to be in $H.P.$ (Harmonic Progression).

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

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