Question

If \[a\] , $ b $ , $ c $ are in AP, then $ \dfrac{1}{{\sqrt a + \sqrt b }} $ , $ \dfrac{1}{{\sqrt a + \sqrt c }} $ , $ \dfrac{1}{{\sqrt b + \sqrt c }} $ are in
1. AP
2. GP
3. HP
4. None of these

Answer 1

Hint: According to the definition of Arithmetic Progression of a series,
Arithmetic Progression is a sequence of terms in which the difference between any two consecutive terms is always the same.
For example, if $ x $ , $ y $ , $ z $ are in AP, then
 $ y - x = z - y $
Interchanging the terms we get,
 $ \Rightarrow 2y = x + z $
Therefore, if the given sequence is in Arithmetic Progression, then it should satisfy the condition that
 $ 2y = x + z $

Complete step-by-step answer:
It is given that $ a $ , $ b $ , $ c $ are in Arithmetic Progression.
We know that if the sequence $ a $ , $ b $ , $ c $ are in AP, then
 $ 2b = a + c - - - - - - \left( 1 \right) $
First let us assume that the sequence given $ \dfrac{1}{{\sqrt a + \sqrt b }} $ , $ \dfrac{1}{{\sqrt a + \sqrt c }} $ , $ \dfrac{1}{{\sqrt b + \sqrt c }} $ is in AP.
Therefore, we have to prove that the terms $ \dfrac{1}{{\sqrt a + \sqrt b }} $ , $ \dfrac{1}{{\sqrt a + \sqrt c }} $ , $ \dfrac{1}{{\sqrt b + \sqrt c }} $ satisfies the condition that
 $ 2\left( {\dfrac{1}{{\sqrt a + \sqrt c }}} \right) = \dfrac{1}{{\sqrt a + \sqrt b }} + \dfrac{1}{{\sqrt b + \sqrt c }} $
Taking L.C.M on the right-hand side we get,
 $ \dfrac{2}{{\sqrt a + \sqrt c }} = \dfrac{{\sqrt b + \sqrt c + \sqrt a + \sqrt b }}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt b + \sqrt c } \right)}} $
Multiplying terms in denominator of R.H.S we get,
 $ \Rightarrow \dfrac{2}{{\sqrt a + \sqrt c }} = \dfrac{{2\sqrt b + \sqrt c + \sqrt a }}{{\sqrt {ab} + \sqrt {ac} + b + \sqrt {bc} }} $
On cross multiplication, we get
 $ 2\sqrt {ab} + 2\sqrt {ac} + 2b + 2\sqrt {bc} = 2\sqrt {ab} + 2\sqrt {bc} + a + c + 2\sqrt {ac} $
On cancelling some terms present on both sides, we get
 $ 2b = a + c $
On substituting equation $ \left( 1 \right) $ in above equation, we get
 $ a + c = a + c $
Therefore L.H.S is equal to R.H.S.
This satisfies the above condition of arithmetic progression.
Hence, we have proved that the terms given $ \dfrac{1}{{\sqrt a + \sqrt b }} $ , $ \dfrac{1}{{\sqrt a + \sqrt c }} $ , $ \dfrac{1}{{\sqrt b + \sqrt c }} $ are in Arithmetic Progression.
So, the correct answer is “Option 1”.

Note: If the above condition does not satisfy, then the terms are not in arithmetic progression.
Then we have to know if the terms are in geometric progression or harmonic progression.
If the terms $ a $ , $ b $ , $ c $ are in geometric progression then the terms should satisfy the equation,
 $ {b^2} = ac $
If the terms $ a $ , $ b $ , $ c $ are in harmonic progression then the terms should satisfy the equation,
 $ b = \dfrac{{2ac}}{{a + c}} $
If the above equation does not satisfy then the answer should be none of these.