| Consider the following statements: |
| 1. \[\frac{1}{ab},\frac{1}{ca}\] and \[\frac{1}{bc}\] are in A.P. |
| 2. \[\frac{1}{\sqrt{b}+\sqrt{c}},\frac{1}{\sqrt{c}+\sqrt{a}}\] and \[\frac{1}{\sqrt{a}+\sqrt{b}}\] are in A.P. |
| Which of the statements given above is/are correct? |
A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 and 2
Correct Answer: C
Solution :
| [c] Let \[\frac{1}{ab},\frac{1}{ca},\frac{1}{bc}\] are in AP. |
| \[\Rightarrow \,\,\,\frac{1}{ca}-\frac{1}{ab}=\frac{1}{bc}-\frac{1}{ca}\] |
| \[\Rightarrow \,\,\,\,\frac{1}{a}\left( \frac{1}{c}-\frac{1}{b} \right)=\frac{1}{c}\left( \frac{1}{b}-\frac{1}{a} \right)\] |
| \[\Rightarrow \,\,\,\,\frac{b-c}{abc}=\frac{a-b}{abc}\] |
| \[\Rightarrow \,\,\,\,b-c=a-b\Rightarrow 2b=a+c\] |
| \[\Rightarrow \] a, b, c are in AP. Which is true |
| Now, \[\frac{1}{\sqrt{b}+\sqrt{c}},\frac{1}{\sqrt{c}+\sqrt{a}},\frac{1}{\sqrt{a}+\sqrt{b}}\] are in A.P. |
| \[\therefore \,\,\,\,\,\frac{2}{\sqrt{c}+\sqrt{a}}=\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{a}+\sqrt{b}}\] |
| \[\Rightarrow \,\,\,2\left( \sqrt{b}+\sqrt{c} \right)\left( \sqrt{a}+\sqrt{b} \right)\] |
| \[=\left( \sqrt{c}+\sqrt{a} \right)\left( \sqrt{a}+2\sqrt{b}+\sqrt{c} \right)\] |
| \[\Rightarrow \,\,\,2\left( \sqrt{ab}+b+\sqrt{ac}+\sqrt{bc} \right)\] |
| \[=\sqrt{ac}+2\sqrt{bc}+c+a+2\sqrt{ab}+\sqrt{ac}\] |
| \[\Rightarrow \,\,2\sqrt{ab}+2b+2\sqrt{ac}+2\sqrt{bc}\] |
| \[=2\sqrt{ac}+2\sqrt{bc}+2\sqrt{ab}+c+a\] |
| \[\Rightarrow \,\,\,\,2b=a+c\] |
| \[\Rightarrow \] a, b, c are in A.P. Which is true. |
| Hence, both the statements are correct. |
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