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.

Answer 1

Given: $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
$\therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow 2ac=ab+bc\dots \dots \left(i\right)$
(i) To prove: $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P.
$2\left(\frac{a+c}{b}\right)=\frac{b+c}{a}+\frac{a+b}{c}\Rightarrow 2ac(a+c)=bc(b+c)+ab(a+b)$
LHS: $2ac(a+c)$
$=(ab+bc)(a+c)$ [From (i)]
$=a^{2}b+2abc+b{c}^2$
RHS: $bc(b+c)+ab(a+c)$
$=b^{2}c+b{c}^2+a^{2}b+a{b}^2=b^{2}c+a{b}^2+b{c}^2+a^{2}b=b(bc+ab)+b{c}^2+a^{2}b=2abc+b{c}^2+a^{2}b$
$=a^{2}b+2abc+b{c}^2$ [From (i)]
$\therefore$ LHS = RHS
Hence proved.
(ii) $a(b+c),b(c+a),c(a+b)$ are in A.P.
if $b(c+a)-a(b+c)=c(a+b)-b(c+a)$
LHS $=b(c+a)-a(b+c)$
$=bc+ab-ab-ac=c(b-a)\dots \dots \left(i\right)$
RHS $=c(a+b)-b(c+a)=ca+cb-bc-ba=a(c-b)ldots\dots \left(ii\right)$
and $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
$\therefore \frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{c}$
Or $c(b-a)=a(c-b)\dots \dots \left(iii\right)$
From (i), (ii) and (iii)
$a(b+c),b(c+a),c(a+b)$ are in A.P.