If a, b, c in mathbb{R}^+ are such that 2a, b | vedclass

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(C) Given that $2a, b, 4c$ are in $A.P.$,we have $2b = 2a + 4c$,which simplifies to $b = a + 2c$.Given that $c, a, b$ are in $G.P.$,we have $a^2 = bc$

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$c, a, a + 2c$ are in $G.P.$

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(C) Given that $2a, b, 4c$ are in $A.P.$,we have $2b = 2a + 4c$,which simplifies to $b = a + 2c$.Given that $c, a, b$ are in $G.P.$,we have $a^2 = bc$.Substituting $b = a + 2c$ into the second equation: $a^2 = (a + 2c)c = ac + 2c^2$.Rearranging gives $a^2 - ac - 2c^2 = 0$.Factoring the quadratic: $(a - 2c)(a + c) = 0$.Since $a, c \in \mathbb{R}^+$,we must have $a = 2c$.Now,check the options:For option $B$: The sequence is $c, a, a + 2c$. Substituting $a = 2c$,we get $c, 2c, 2c + 2c$,which is $c, 2c, 4c$.This is a $G.P.$ with common ratio $2$,not an $A.P.$For option $C$: The sequence is $c, a, a + 2c$. Substituting $a = 2c$,we get $c, 2c, 4c$. This is a $G.P.$ with common ratio $2$.Thus,$c, a, a + 2c$ are in $G.P.$

If $a, b, c \in \mathbb{R}^+$ are such that $2a, b, 4c$ are in $A.P.$ and $c, a, b$ are in $G.P.$,then:

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