Sequence And Series Question 411
Question: Let a, b, c be in AP.
Consider the following statements:
- $ \frac{1}{ab},\frac{1}{ca} $ and $ \frac{1}{bc} $ are in A.P.
- $ \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?
Options:
A) 1 only
B) 2 only
C) Both 1 and 2
D) Neither 1 and 2
Show Answer
Answer:
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}( \frac{1}{c}-\frac{1}{b} )=\frac{1}{c}( \frac{1}{b}-\frac{1}{a} ) $
$ \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( \sqrt{b}+\sqrt{c} )( \sqrt{a}+\sqrt{b} ) $ $ =( \sqrt{c}+\sqrt{a} )( \sqrt{a}+2\sqrt{b}+\sqrt{c} ) $
$ \Rightarrow ,2( \sqrt{ab}+b+\sqrt{ac}+\sqrt{bc} ) $ $ =\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.
BETA
