Let $$a, b, c$$  be in A.P. Consider the following statements :
$$\eqalign{ & 1.\,\,\,\frac{1}{{ab}},\frac{1}{{ca}}{\text{and}}\frac{1}{{bc}}{\text{are in A}}{\text{.P}}{\text{.}} \cr & {\text{2}}{\text{.}}\,\,\,\frac{1}{{\sqrt b + \sqrt c }},\frac{1}{{\sqrt c + \sqrt a }}{\text{and}}\frac{1}{{\sqrt a + \sqrt b }}{\text{are in A}}{\text{.P}}{\text{.}} \cr} $$
Which of the statements given above is/are correct ?

Solution :
$$\eqalign{ & {\text{Let, }}\frac{1}{{ab}},\frac{1}{{ca}},\frac{1}{{bc}}{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \frac{1}{{ca}} - \frac{1}{{ab}} = \frac{1}{{bc}} - \frac{1}{{ca}} \cr & \Rightarrow \frac{1}{a}\left( {\frac{1}{c} - \frac{1}{b}} \right) = \frac{1}{c}\left( {\frac{1}{b} - \frac{1}{a}} \right) \cr & \Rightarrow \frac{{b - c}}{{abc}} = \frac{{a - b}}{{abc}} \cr & \Rightarrow b - c = a - b \cr & \Rightarrow 2b = a + c \cr} $$
⇒ $$a, b, c$$  are in A.P. Which is true
$$\eqalign{ & {\text{Now, }}\frac{1}{{\sqrt b + \sqrt c }},\frac{1}{{\sqrt c + \sqrt a }},\frac{1}{{\sqrt a + \sqrt b }}{\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore \frac{2}{{\sqrt c + \sqrt a }} = \frac{1}{{\sqrt b + \sqrt c }} + \frac{1}{{\sqrt a + \sqrt b }} \cr & \Rightarrow 2\left( {\sqrt b + \sqrt c } \right)\left( {\sqrt a + \sqrt b } \right) = \left( {\sqrt c + \sqrt a } \right)\left( {\sqrt a + 2\sqrt b + \sqrt c } \right) \cr & \Rightarrow 2\left( {\sqrt {ab} + b + \sqrt {ac} + \sqrt {bc} } \right) = \sqrt {ac} + 2\sqrt {bc} + c + a + 2\sqrt {ab} + \sqrt {ac} \cr & \Rightarrow 2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = 2\sqrt {ac} + 2\sqrt {bc} + 2\sqrt {ab} + c + a \cr & \Rightarrow 2b = a + c \cr} $$
⇒ $$a, b, c$$  are in A.P. Which is true.
Hence, both the statements are correct