If (b+c), (c+a), (a+b) are in H.P.,then a^2, | vedclass

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(A) Given that $(b+c), (c+a), (a+b)$ are in $H.P.$Therefore,their reciprocals $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in $A.P.$This implies:

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$A.P.$

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(A) Given that $(b+c), (c+a), (a+b)$ are in $H.P.$Therefore,their reciprocals $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in $A.P.$This implies: $\frac{2}{c+a} = \frac{1}{b+c} + \frac{1}{a+b}$$\frac{2}{c+a} = \frac{a+b+b+c}{(b+c)(a+b)} = \frac{a+2b+c}{ab+b^2+ac+bc}$$2(ab+b^2+ac+bc) = (c+a)(a+2b+c)$$2ab+2b^2+2ac+2bc = ac+2bc+c^2+a^2+2ab+ac$$2ab+2b^2+2ac+2bc = a^2+c^2+2ac+2ab+2bc$Subtracting $(2ab+2ac+2bc)$ from both sides,we get:$2b^2 = a^2+c^2$Since $2b^2 = a^2+c^2$,it follows that $a^2, b^2, c^2$ are in $A.P.$

If $(b+c), (c+a), (a+b)$ are in $H.P.$,then $a^2, b^2, c^2$ are in.......

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