If the angles A, B, C of a triangle are in A. | vedclass

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(A) Given that $A, B, C$ are in $A.P.$,we have $A + C = 2B$. Since $A + B + C = 180^\circ$,we get $3B = 180^\circ$,so $B = 60^\circ$.Given that $a, b,

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$A.P.$

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(A) Given that $A, B, C$ are in $A.P.$,we have $A + C = 2B$. Since $A + B + C = 180^\circ$,we get $3B = 180^\circ$,so $B = 60^\circ$.Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Substituting $B = 60^\circ$ and $ac = b^2$: $\cos 60^\circ = \frac{a^2 + c^2 - b^2}{2b^2}$.$\frac{1}{2} = \frac{a^2 + c^2 - b^2}{2b^2} \Rightarrow b^2 = a^2 + c^2 - b^2$.Therefore,$a^2 + c^2 = 2b^2$,which implies that $a^2, b^2, c^2$ are in $A.P.$

If the angles $A, B, C$ of a triangle are in $A.P.$ and the sides $a, b, c$ opposite to these angles are in $G.P.$,then $a^2, b^2, c^2$ are in

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