Question

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

Answer 1

Given $\frac{1}{a},\frac{1}{b},\frac{1}{c}areinAP$

(i)

$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

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

$ac-bc=ab-ac$

$ac+ac=ab+bc$

$2ac=ab+bc$

$2ca=ab+bc$

$Thusbc,caandabareinAP$

(ii)

We know that if, b(c + a) – a(b + c) = c(a + b) – b(c + a)

Consider LHS: b(c + a) – a(b + c)
Upon simplification we get,
b(c + a) – a(b + c) = bc + ba – ab – ac
= c (b - a)

Now, c(a + b) – b(c + a) = ca + cb – bc – ba
= a (c-b)

We know, $\frac{1}{a},\frac{1}{b},\frac{1}{c}areinAP$

So, $\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$
Or c(b - a) = a(c - b)

Hence, given terms are in AP.