If x, y, z are three positive numbers in G.P. | vedclass

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(D) Since $x, y, z$ are in $G.P.$,we have $y^2 = xz$.Taking the natural logarithm on both sides,$2 \ln y = \ln x + \ln z$,which implies $\ln x, \ln y,

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$A.G.P.$

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(D) Since $x, y, z$ are in $G.P.$,we have $y^2 = xz$.Taking the natural logarithm on both sides,$2 \ln y = \ln x + \ln z$,which implies $\ln x, \ln y, \ln z$ are in $A.P.$Let $a = \ln x, b = \ln y, c = \ln z$. Then $a, b, c$ are in $A.P.$We are given the terms $T_1 = \frac{1+a}{2}, T_2 = \frac{1+b}{4}, T_3 = \frac{1+c}{8}$.Since $a, b, c$ are in $A.P.$,let $a = A-d, b = A, c = A+d$.Then $T_1 = \frac{1+A-d}{2}, T_2 = \frac{1+A}{4}, T_3 = \frac{1+A+d}{8}$.Checking for $G.P.$: $T_2^2 = \frac{(1+A)^2}{16}$ and $T_1 	imes T_3 = \frac{(1+A)^2 - d^2}{16}$. Since $T_2^2 
eq T_1 T_3$,they are not in $G.P.$Checking for $A.P.$: $T_2 - T_1 = \frac{1+A}{4} - \frac{1+A-d}{2} = \frac{1+A - 2 - 2A + 2d}{4} = \frac{-1-A+2d}{4}$.$T_3 - T_2 = \frac{1+A+d}{8} - \frac{1+A}{4} = \frac{1+A+d - 2 - 2A}{8} = \frac{-1-A+d}{8}$.Since $T_2 - T_1 
eq T_3 - T_2$,they are not in $A.P.$However,the sequence is of the form $\frac{a_n}{2^n}$ where $a_n$ is in $A.P.$,which defines an $A.G.P.$

If $x, y, z$ are three positive numbers in $G.P.$,then $\frac{1 + \ln x}{2}, \frac{1 + \ln y}{4}, \frac{1 + \ln z}{8}$ are in

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