Question

If \[{a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }} \ldots ..\] are in A.P. where \[{a_i} > {\text{ }}0\]for all show that
 $ \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .........\dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = \dfrac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }} $

Answer 1

Hint: To solve this question, we will take L.H.S. first, where we start with rationalizing as it contains imaginary numbers.
Then using the formulas of A.P. we will show the L.H.S. equals to R.H.S.

Complete step-by-step answer:
Step 1: We have been given, \[{a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }} \ldots ..\] in A.P. where \[{a_i} > {\text{ }}0\]
And we have to show that, $ \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .........\dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = \dfrac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }} $
Now let us take the L.H.S. of the given equation, $ \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .........\dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} $
On rationalizing the above L.H.S. of the given equation, we get
 $ \begin{gathered}
   = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{{a_1} - {a_2}}} + \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{{a_2} - {a_3}}} + .........\dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{{a_{n - 1}} - {a_n}}} \\
    \\
\end{gathered} $ \[ \ldots \ldots \ldots .{\text{ }}eq.{\text{ }}\left( 1 \right)\]
Step 2: So, the formula from AP which we used now is, \[{a_2}-{\text{ }}{a_1} = {\text{ }}d,\]
where, a2 \[ = \] second term of AP
 a1 \[ = \] first term of AP
d \[ = \] difference between two consecutive terms in AP
On substituting \[{a_2}-{\text{ }}{a_1} = {\text{ }}d,\] in \[eq.{\text{ }}\left( 1 \right),\]we get
 $ = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{ - d}} + \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{ - d}} + .........\dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{ - d}} $
 $ = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} + \sqrt {{a_2}} - \sqrt {{a_3}} ....... + \sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{ - d}} $
 $ = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_n}} }}{{ - d}} $
Now, on again rationalizing, we get
 $ \begin{gathered}
   = \dfrac{{{a_1} - {a_n}}}{{ - d\sqrt {{a_1}} + \sqrt {{a_n}} }} \\
    \\
\end{gathered} $ \[ \ldots \ldots \ldots \ldots \ldots \ldots \ldots .{\text{ }}eq.{\text{ }}\left( 2 \right)\]
Step 3: Now, put the value of \[{a_n} = {\text{ }}{a_1} + {\text{ }}\left( {n - 1} \right)d,\]
where, a $ = $ first term of AP
n $ = $ number of terms in AP
d $ = $ difference between two consecutive terms in AP.
On substituting \[{a_2}-{\text{ }}{a_1} = {\text{ }}d,\] in \[eq.{\text{ }}\left( 2 \right),\]we get

 $ \begin{gathered}
   = \dfrac{{ - (n - 1)d}}{{ - d\sqrt {{a_1}} + \sqrt {{a_n}} }} \\
    \\
\end{gathered} $
 $ \begin{gathered}
   = \dfrac{{(n - 1)}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }} \\
    \\
\end{gathered} $
which is equal to R.H.S.
Thus, $ \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .........\dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = \dfrac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }} $

Note: In rationalization, numerator and denominator is multiplied by same root number. It is done to remove the imaginary number from the given equation.
For ex: $ \dfrac{3}{{2 + \sqrt 2 }} = \dfrac{3}{{2 + \sqrt 2 }} \times \dfrac{{2 - \sqrt 2 }}{{2 - \sqrt 2 }} = \dfrac{{3(2 - \sqrt {2)} }}{{4 - 2}} = \dfrac{{6 - 3\sqrt 2 }}{2} $