Question

if a,b,c are in AP then prove that
$\frac{1}{ab}, \frac{1}{bc} ,\frac{1}{ca}$
are in AP​

Answer 1

Step-by-step explanation:

Given,

a, b and c are in AP

Therefore,

from theory of AP,

b + c = 2a ......(i)

Now,

we have,

$\frac{1}{ab}, \frac{1}{bc} \:\:and\:\:\frac{1}{ca}$

Adding the first and last terms,

we get,

$\frac{1}{ab} + \frac{1}{ac} \\ \\ = \frac{1}{a} ( \frac{1}{b} + \frac{1}{c} ) \\ \\ = \frac{1}{a} \times \frac{b + c}{bc} \\ \\ = \frac{b + c}{abc}$

But,

from eqn

b + c = 2a

Therefore,

putting the values,

we get,

$= \frac{2a}{abc} \\ \\ = \frac{2}{bc}$

Thus, we get,

$\frac{1}{ab} + \frac{1}{ca} = \frac{2}{bc}$

Hence,

$\bold{ \frac{1}{ab}, \frac{1}{bc} ,\frac{1}{ca}}$ are in AP

Thus

Proved