If {a^2}, {b^2}, {c^2} are in A.P.,then frac{ | vedclass

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(A) Given that ${a^2}, {b^2}, {c^2}$ are in $A.P.$This implies ${b^2} - {a^2} = {c^2} - {b^2}$.Adding ${ab + bc + ca}$ to both sides or manipulating t

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$A.P.$

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(A) Given that ${a^2}, {b^2}, {c^2}$ are in $A.P.$This implies ${b^2} - {a^2} = {c^2} - {b^2}$.Adding ${ab + bc + ca}$ to both sides or manipulating the terms,we observe that if we add ${ab + bc + ca}$ to each term of the sequence,the property of $A.P.$ is preserved.Alternatively,consider the terms $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$.Add $1$ to each term: $\frac{a}{b+c} + 1 = \frac{a+b+c}{b+c}$,$\frac{b}{c+a} + 1 = \frac{a+b+c}{c+a}$,$\frac{c}{a+b} + 1 = \frac{a+b+c}{a+b}$.Since ${a^2}, {b^2}, {c^2}$ are in $A.P.$,it follows that ${b+c}, {c+a}, {a+b}$ are in $H.P.$ (or their reciprocals are in $A.P.$).Therefore,$\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in $A.P.$

If ${a^2}, {b^2}, {c^2}$ are in $A.P.$,then $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ will be in

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