Question

If $a(\frac{1}{b}+\frac{1}{c}),b(\frac{1}{c}+\frac{1}{a}),c(\frac{1}{a}+\frac{1}{b})$ are in A.P. prove that $a,b,c$ are in A.P.

Answer 1

$\text{It is given that},a(\frac{1}{b}+\frac{1}{c}),b(\frac{1}{c}+\frac{1}{a}),c(\frac{1}{a}+\frac{1}{b})\text{are in}A.P.$

$\therefore b(\frac{1}{c}+\frac{1}{a})-a(\frac{1}{b}+\frac{1}{c})=c(\frac{1}{a}+\frac{1}{b})-b(\frac{1}{c}+\frac{1}{a})$

$\Rightarrow \frac{b(a+c)}{ac}-\frac{a(b+c)}{bc}=\frac{c(a+b)}{ab}-\frac{b(a+c)}{ac}$

$\Rightarrow \frac{b^{2}a+b^2-c-a^{2}b-a^{2}c}{abc}=\frac{c^{2}a+c^{2}b-b^{2}a-b^{2}c}{abc}$

$\Rightarrow b^{2}a-a^{2}b+b^{2}c-a^{2}c=c^{2}a-b^{2}a+c^{2}b-b^{2}c$

$\Rightarrow ab(b-a)+c(b^2-a^2)=a(c^2-b^2)+bc(c-b)$

$\Rightarrow ab(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+bc(c-b)$

$\Rightarrow b-a=c-b$

$\text{Thus a, b and c are in A.P.}$