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) $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Since $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P., we have:
$\frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b}\Rightarrow \frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc}\Rightarrow \frac{(a+b)(a-b)+c(a-b)}{ab}=\frac{(b+c)(b-c)+a(b-c)}{bc}\Rightarrow \frac{(a-b)(a+b+c)}{ab}=\frac{(b-c)(a+b+c)}{bc}\Rightarrow \frac{(a-b)}{ab}=\frac{(b-c)}{bc}\Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$
Hence, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

(ii) bc, ca, ab are in A.P.
$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}\Rightarrow \frac{(a-b)}{ab}=\frac{(b-c)}{bc}\Rightarrow \frac{(a-b)}{a}=\frac{(b-c)}{c}\Rightarrow (a-b)c=a(b-c)\Rightarrow ac-bc=ab-ac$
Hence, bc, ca, ab are in A.P.