If a, b, c are in H.P.,then frac{a}{b + c}, f | vedclass

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(C) Given that $a, b, c$ are in $H.P.$This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Divide each term by the sum $(a+b+c)$ or

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$H.P.$

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(C) Given that $a, b, c$ are in $H.P.$This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$Divide each term by the sum $(a+b+c)$ or multiply by $(a+b+c)$,we get $\frac{a+b+c}{a}, \frac{a+b+c}{b}, \frac{a+b+c}{c}$ are in $A.P.$Subtracting $1$ from each term: $\frac{a+b+c}{a} - 1, \frac{a+b+c}{b} - 1, \frac{a+b+c}{c} - 1$ are in $A.P.$This simplifies to $\frac{b+c}{a}, \frac{a+c}{b}, \frac{a+b}{c}$ are in $A.P.$Taking the reciprocal of each term,we get $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in $H.P.$

If $a, b, c$ are in $H.P.$,then $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in

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