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

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

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

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(C) Given $2a, b, 4c$ are in $A.P.$,we have $2b = 2a + 4c$,which simplifies to $b = a + 2c$.Given $c, a, b$ are in $G.P.$,we have $a^2 = bc$.Substituting $b = a + 2c$ into $a^2 = bc$,we get $a^2 = c(a + 2c) = ac + 2c^2$.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 terms are $c, a, a+2c$. Substituting $a=2c$,we get $c, 2c, 2c+2c = c, 2c, 4c$.These are in $G.P.$ with common ratio $2$,but let's check if they are in $A.P.$: $2c - c = c$ and $4c - 2c = 2c$. Since $c 
eq 2c$,they are not in $A.P.$For option $C$: The terms are $c, a, a+2c$. Substituting $a=2c$,we get $c, 2c, 4c$. These are in $G.P.$ with common ratio $2$.

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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