If alpha, beta, gamma are the geometric means | vedclass

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(A) Given that $\alpha, \beta, \gamma$ are geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively.Therefore,$\alpha^2 = (ca)(ab) = a^2b

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$A.P.$

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(A) Given that $\alpha, \beta, \gamma$ are geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively.Therefore,$\alpha^2 = (ca)(ab) = a^2bc$,$\beta^2 = (ab)(bc) = b^2ca$,and $\gamma^2 = (bc)(ca) = c^2ab$.Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.We need to check if $\alpha^2, \beta^2, \gamma^2$ are in $A.P.$,which requires $2\beta^2 = \alpha^2 + \gamma^2$.Substituting the values: $2(b^2ca) = a^2bc + c^2ab$.Dividing both sides by $abc$ (assuming $a, b, c 
eq 0$),we get $2b^2 = ac + ac$ is incorrect; rather,$2b^2ca = abc(a + c)$.Dividing by $abc$,we get $2b = a + c$,which is true as $a, b, c$ are in $A.P$.Thus,$\alpha^2, \beta^2, \gamma^2$ are in $A.P.$

If $\alpha, \beta, \gamma$ are the geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively,where $a, b, c$ are in $A.P.$,then $\alpha^2, \beta^2, \gamma^2$ are in

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