If a^2, b^2, c^2 are in A.P.,then (b + c)^{-1 | vedclass

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(C) Given that $a^2, b^2, c^2$ are in $A.P.$Adding $(ab + bc + ca)$ to each term,we get:$a^2 + ab + bc + ca, b^2 + ab + bc + ca, c^2 + ab + bc + ca$ a

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$A.P.$

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(C) Given that $a^2, b^2, c^2$ are in $A.P.$Adding $(ab + bc + ca)$ to each term,we get:$a^2 + ab + bc + ca, b^2 + ab + bc + ca, c^2 + ab + bc + ca$ are in $A.P.$Factoring the expressions:$a(a + b) + c(a + b), b(b + a) + c(b + a), c(c + b) + a(c + b)$ are in $A.P.$This simplifies to:$(a + b)(a + c), (b + a)(b + c), (c + a)(c + b)$ are in $A.P.$Dividing each term by $(a + b)(b + c)(c + a)$,we get:$\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in $A.P.$Thus,$(b + c)^{-1}, (c + a)^{-1}, (a + b)^{-1}$ are in $A.P.$

If $a^2, b^2, c^2$ are in $A.P.$,then $(b + c)^{-1}, (c + a)^{-1}$ and $(a + b)^{-1}$ will be in

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