Question

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

(i) $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

(ii) bc, ca, ab are in A.P.

Answer 1

(i) Since $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P., we have:

$$\begin{gathered} \frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b} \\ \Rightarrow \frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc} \\ \Rightarrow \frac{\left(a+b\right)\left(a-b\right)+c\left(a-b\right)}{ab}=\frac{\left(b+c\right)\left(b-c\right)+a\left(b-c\right)}{bc} \\ \Rightarrow \frac{\left(a-b\right)\left(a+b+c\right)}{ab}=\frac{\left(b-c\right)\left(a+b+c\right)}{bc} \\ \Rightarrow \frac{\left(a-b\right)}{ab}=\frac{\left(b-c\right)}{bc} \\ \Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b} \end{gathered}$$

Hence, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.


(ii) Since $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., we have:

$$\begin{gathered} \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b} \\ \Rightarrow \frac{\left(a-b\right)}{ab}=\frac{\left(b-c\right)}{bc} \\ \Rightarrow \frac{\left(a-b\right)}{a}=\frac{\left(b-c\right)}{c} \\ \Rightarrow \left(a-b\right)c=a\left(b-c\right) \\ \Rightarrow ac-bc=ab-ac \end{gathered}$$

Hence, bc, ca, ab are in A.P.