# Solve the following

Question:

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

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

(ii) a (b +c), b (c + a), c (a +b) are in A.P.

Solution:

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

$\therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\Rightarrow 2 a c=a b+b c \quad \ldots(1)$

(i) To prove: $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P.

$2\left(\frac{a+c}{b}\right)=\frac{b+c}{a}+\frac{a+b}{c}$

$\Rightarrow 2 a c(a+c)=b c(b+c)+a b(a+b)$

$\mathrm{LHS}: 2 a c(a+c)$

$=(a b+b c)(a+c) \quad(\operatorname{From}(1))$

$=a^{2} b+2 a b c+b c^{2}$

RHS : $b c(b+c)+a b(a+b)$

$=b^{2} c+b c^{2}+a^{2} b+a b^{2}$

$=b^{2} c+a b^{2}+b c^{2}+a^{2} b$

$=b(b c+a b)+b c^{2}+a^{2} b$

$=2 a b c+b c^{2}+a^{2} b$

$=a^{2} b+2 a b c+b c^{2} \quad(\operatorname{From}(1))$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.

(ii) To prove: $a(b+c), b(c+a), c(a+b)$ are in A.P.

$\Rightarrow 2 b(c+a)=a(b+c)+c(a+b)$

LHS : $2 b(c+a)$

$=2 b c+2 b a$

RHS : $a(b+c)+c(a+b)$

$=a b+a c+a c+b c$

$=a b+2 a c+b c$

$=a b+a b+b c+b c$      (From (1))

$=2 a b+2 b c$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.