If frac{a}{b}, frac{b}{c}, frac{c}{a} are in | vedclass

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(A) Given that $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in $H.P.$Therefore,their reciprocals $\frac{b}{a}, \frac{c}{b}, \frac{a}{c}$ are in $A.P.$T

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$a^2b, c^2a, b^2c$ are in $A.P.$

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(A) Given that $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in $H.P.$Therefore,their reciprocals $\frac{b}{a}, \frac{c}{b}, \frac{a}{c}$ are in $A.P.$This implies $2 	imes (\frac{c}{b}) = \frac{b}{a} + \frac{a}{c}$.Multiplying both sides by $abc$,we get $2ac^2 = b^2c + a^2b$.This equation represents the condition for $a^2b, c^2a, b^2c$ to be in $A.P.$ because $2(c^2a) = a^2b + b^2c$ is equivalent to $2c^2a = a^2b + b^2c$.

If $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in $H.P.$,then

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