
Let a, b, c be in AP.
Consider the following statements:
1. \[\dfrac{1}{{ab}},\dfrac{1}{{ca}}\]and \[\dfrac{1}{{bc}}\]are in AP
2. \[\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }}\] and \[\dfrac{1}{{\sqrt a + \sqrt b }}\] are in AP
Which of the statements given above is/are correct?
A) \[1\]Only
B) \[2\]Only
C) Both \[1\]and \[2\]
D) Neither \[1\]and \[2\]
Answer
161.4k+ views
Hint: in this question we have to find which statement among the two is true. In order to find this we have to use the property of AP according to which if three terms are in AP then twice of the middle term is equal to the sum of first term and third term. Now apply this concept on given statement if after simplification we get \[2b = a + c\] then statement is true otherwise statement is false.
Formula Used: If a, b, c are in AP then
\[2b = a + c\]
Complete step by step solution: Given: a, b, c three terms which are following the AP series
Now we have
\[\dfrac{1}{{ab}},\dfrac{1}{{ca}}\]and \[\dfrac{1}{{bc}}\]are in AP
In AP difference is common among the terms
\[\dfrac{1}{{ca}} - \dfrac{1}{{ab}} = \dfrac{1}{{bc}} - \dfrac{1}{{ca}}\]
\[\dfrac{1}{a}(\dfrac{1}{c} - \dfrac{1}{b}) = \dfrac{1}{c}(\dfrac{1}{b} - \dfrac{1}{a})\]
\[\dfrac{1}{a}(\dfrac{{b - c}}{{cb}}) = \dfrac{1}{c}(\dfrac{{a - b}}{{ab}})\]
\[(\dfrac{{b - c}}{{acb}}) = (\dfrac{{a - b}}{{abc}})\]
On simplification we get
\[b - c = a - b\]
\[2b = a + c\]
Hence statement first is true
Now we have
\[\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }}\] and \[\dfrac{1}{{\sqrt a + \sqrt b }}\] are in AP
In AP difference is common among the terms
\[\dfrac{1}{{\sqrt c + \sqrt a }} - \dfrac{1}{{\sqrt b + \sqrt c }} = \dfrac{1}{{\sqrt a + \sqrt b }} - \dfrac{1}{{\sqrt c + \sqrt a }}\]
\[\dfrac{2}{{\sqrt c + \sqrt a }} = \dfrac{1}{{\sqrt b + \sqrt c }} + \dfrac{1}{{\sqrt a + \sqrt b }}\]
\[2(\sqrt b + \sqrt c )(\sqrt a + \sqrt b ) = (\sqrt c + \sqrt a )(\sqrt a + 2\sqrt b + \sqrt c )\]
\[2(\sqrt {ab} + b + \sqrt {ac} + \sqrt {bc} ) = \sqrt {ac} + 2\sqrt {bc} + c + a + 2\sqrt {ab} + \sqrt {ac} \]
\[2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = 2\sqrt {ac} + 2\sqrt {bc} + c + a + 2\sqrt {ab} + 2\sqrt {ac} \]
On simplification we get
\[2b = a + c\]
Hence statement second is true
Option ‘C’ is correct
Note: Here we must remember that if three terms are in AP then twice of the middle term is equal to the sum of first term and third term.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Formula Used: If a, b, c are in AP then
\[2b = a + c\]
Complete step by step solution: Given: a, b, c three terms which are following the AP series
Now we have
\[\dfrac{1}{{ab}},\dfrac{1}{{ca}}\]and \[\dfrac{1}{{bc}}\]are in AP
In AP difference is common among the terms
\[\dfrac{1}{{ca}} - \dfrac{1}{{ab}} = \dfrac{1}{{bc}} - \dfrac{1}{{ca}}\]
\[\dfrac{1}{a}(\dfrac{1}{c} - \dfrac{1}{b}) = \dfrac{1}{c}(\dfrac{1}{b} - \dfrac{1}{a})\]
\[\dfrac{1}{a}(\dfrac{{b - c}}{{cb}}) = \dfrac{1}{c}(\dfrac{{a - b}}{{ab}})\]
\[(\dfrac{{b - c}}{{acb}}) = (\dfrac{{a - b}}{{abc}})\]
On simplification we get
\[b - c = a - b\]
\[2b = a + c\]
Hence statement first is true
Now we have
\[\dfrac{1}{{\sqrt b + \sqrt c }},\dfrac{1}{{\sqrt c + \sqrt a }}\] and \[\dfrac{1}{{\sqrt a + \sqrt b }}\] are in AP
In AP difference is common among the terms
\[\dfrac{1}{{\sqrt c + \sqrt a }} - \dfrac{1}{{\sqrt b + \sqrt c }} = \dfrac{1}{{\sqrt a + \sqrt b }} - \dfrac{1}{{\sqrt c + \sqrt a }}\]
\[\dfrac{2}{{\sqrt c + \sqrt a }} = \dfrac{1}{{\sqrt b + \sqrt c }} + \dfrac{1}{{\sqrt a + \sqrt b }}\]
\[2(\sqrt b + \sqrt c )(\sqrt a + \sqrt b ) = (\sqrt c + \sqrt a )(\sqrt a + 2\sqrt b + \sqrt c )\]
\[2(\sqrt {ab} + b + \sqrt {ac} + \sqrt {bc} ) = \sqrt {ac} + 2\sqrt {bc} + c + a + 2\sqrt {ab} + \sqrt {ac} \]
\[2\sqrt {ab} + 2b + 2\sqrt {ac} + 2\sqrt {bc} = 2\sqrt {ac} + 2\sqrt {bc} + c + a + 2\sqrt {ab} + 2\sqrt {ac} \]
On simplification we get
\[2b = a + c\]
Hence statement second is true
Option ‘C’ is correct
Note: Here we must remember that if three terms are in AP then twice of the middle term is equal to the sum of first term and third term.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
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