If are in A.P., prove that a, b, c are in A.P.

Given \( a(\dfrac{1}{b}+\dfrac{1}{c}), b(\dfrac{1}{c}+\dfrac{1}{a}), c(\dfrac{1}{a}+\dfrac{1}{b}),\) are in A.P.

Therefore,

\( b(\dfrac{1}{c}+\dfrac{1}{a}), -a(\dfrac{1}{b}+\dfrac{1}{c})\)

= \( c(\dfrac{1}{a}+\dfrac{1}{b})-b(\dfrac{1}{c}+\dfrac{1}{a})\)

\( \dfrac{b(a+c)}{ac}-\dfrac{a(b+c)}{bc}\)

= \( \dfrac{c(a+b)}{ab}-\dfrac{b(a+c)}{ac}\)

= \( \dfrac{b^2a+b^2c-a^2b-a^2c}{abc}\)

= \( \dfrac{c^2a+c^2b-b^2a-b^2c}{abc}\)

= \( b^2a+b^2c-a^2b-a^2c\)

= \( c^2a+c^2b-b^2a-b^2c\)

= \( ab(b-a)+c(b^2-a^2)\)

\( =a(c^2-b^2)bc(c-b)\)

\( ab(b-a)+c(b-a)(b+a)\)

\( = a(c-b)(c+b)+bc(c-b)\)

= \( (b-a)(ab+cb+ca)\)

\( =(c-b)(ac+ab+bc)\)

= \( b-a=c-b\)

Therefore, a, b and c are in A.P.