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

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(c) ${a^2},\;{b^2},\;{c^2}$ are in $A.P.$

Therefore ${a^2} + (ab + bc + ca)$, ${b^2} + (ab + bc + ca)$, ${c^2} + (ab + bc + ca)$ will be in $A.P.$

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(c) ${a^2},\;{b^2},\;{c^2}$ are in $A.P.$

Therefore ${a^2} + (ab + bc + ca)$, ${b^2} + (ab + bc + ca)$, ${c^2} + (ab + bc + ca)$ will be in $A.P.$

$ \Rightarrow $ $\{ a(a + b) + c\,(a + b)\} ,\;\{ b(b + a) + c\,(b + a)\} ,\;$

$c(c + b) + a(b + c)$ will be in $A.P.$

==>$(a + b)(a + c),\;(b + a)(b + c),\;(c + a)(c + b)$ will be in $A.P.$

$ \Rightarrow $$\frac{1}{{b + c}},\;\frac{1}{{c + a}},\;\frac{1}{{a + b}}$ will be in $A.P.$

{ Dividing each term by $(a+b)(b+c)(c+a)$ }

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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