Question

If $ {a_1},{a_2},...,{a_n} $ are in arithmetic progression, where $ {a_i} > 0 $ for all \[\;i\] . Then,
 $ \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}} }} = $
A. $ \dfrac{{{n^2}\left( {n + 1} \right)}}{2} $
B. $ \dfrac{{\left( {n - 1} \right)}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }} $
C. $ \dfrac{{n\left( {n - 1} \right)}}{2} $
D. None of these

Answer 1

Hint: An arithmetic sequence or arithmetic progression is defined as a mathematical sequence in which the difference between two consecutive terms is always a constant. An arithmetic progression is abbreviated as A.P. We also need to learn the three important terms, which are as follows.
A common difference $ \left( d \right) $ is the difference between the first two terms.
 $ {n^{th}} $ term \[({a_n})\]
And, Sum of the first $ n $ terms \[({S_n})\]
Formula to be used:
a) The formula to calculate the $ {n^{th}} $ term of the given arithmetic progression is as follows.
  \[{a_n} = a + \left( {n - 1} \right)d\]
Where, $ a $ denotes the first term, $ d $ denotes the common difference, $ n $ is the number of terms, and $ {a_n} $ is the $ {n^{th}} $ term of the given arithmetic progression.
b) We know that \[{a_n} = a + \left( {n - 1} \right)d\]
 \[{a_n} = a + \left( {n - 1} \right)d \Rightarrow a - {a_n} = - \left( {n - 1} \right)d\]
c)) $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $

Complete step by step answer:
We are given that $ {a_1},{a_2},...,{a_n} $ are in arithmetic progression, where $ {a_i} > 0 $ for all \[\;i\] .
Hence, the common difference $ d $ is given by \[{a_2} - {a_1}\; = {a_3} - {a_2} = \ldots . = {a_n} - {a_{n - 1}}\; = d\]
Multiplying by a minus sign, we get
 \[{a_1} - {a_2}\; = {a_2} - {a_3} = \ldots . = {a_{n - 1}} - {a_n}\; = - d\] ………. $ \left( 1 \right) $
Now, we need to calculate $ \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}} }} $
We shall rationalize the denominator of the above expression.
 \[\dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + ..... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} \times \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{\sqrt {{a_1}} - \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} \times \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{\sqrt {{a_2}} - \sqrt {{a_3}} }} + ..... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} \times \dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}\] \[ = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{\left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\left( {\sqrt {{a_1}} - \sqrt {{a_2}} } \right)}} + \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{\left( {\sqrt {{a_2}} + \sqrt {{a_3}} } \right)\left( {\sqrt {{a_2}} - \sqrt {{a_3}} } \right)}} + ..... + \dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{\left( {\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} } \right)\left( {\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} } \right)}}\]
Here, we shall apply the formula $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $ on the denominators of the right-hand side equation.
 \[\dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + ..... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{{{\left( {\sqrt {{a_1}} } \right)}^2} - {{\left( {\sqrt {{a_2}} } \right)}^2}}} + \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{{{\left( {\sqrt {{a_2}} } \right)}^2} - {{\left( {\sqrt {{a_3}} } \right)}^2}}} + ..... + \dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{{{\left( {\sqrt {{a_{n - 1}}} } \right)}^2} - {{\left( {\sqrt {{a_n}} } \right)}^2}}}\] \[ = \dfrac{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{{a_1} - {a_2}}} + \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{{a_2} - {a_3}}} + ..... + \dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{{a_{n - 1}} - {a_n}}}\]
Now, we shall apply the equation $ \left( 1 \right) $ above.
   $ \Rightarrow \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{{\sqrt {{a_1}} - \sqrt {{a_2}} }}{{ - d}} + \dfrac{{\sqrt {{a_2}} - \sqrt {{a_3}} }}{{ - d}} + ..... + \dfrac{{\sqrt {{a_{n - 1}}} - \sqrt {{a_n}} }}{{ - d}} $
 $ \Rightarrow \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{1}{{ - d}}\left( {\sqrt {{a_1}} - \sqrt {{a_2}} + \sqrt {{a_2}} - \sqrt {{a_3}} + ..... + \sqrt {{a_{n - 1}}} - \sqrt {{a_n}} } \right) $
                                                                                           $ = \dfrac{1}{{ - d}}\left( {\sqrt {{a_1}} - \sqrt {{a_n}} } \right) $
We shall rationalize the numerator of the above expression.
 $ \Rightarrow \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{1}{{ - d}}\left( {\sqrt {{a_1}} - \sqrt {{a_n}} \times \dfrac{{\sqrt {{a_1}} + \sqrt {{a_n}} }}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}} \right) $
Here, we shall apply the formula $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $ on the numerators of the right-hand side equation.
 $ \Rightarrow \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{1}{{ - d}}\left( {\dfrac{{{{\left( {\sqrt {{a_1}} } \right)}^2} - {{\left( {\sqrt {{a_n}} } \right)}^2}}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}} \right) $
                                                                                              $ = \dfrac{1}{{ - d}}\left( {\dfrac{{{a_1} - {a_n}}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}} \right) $ ……… $ \left( 2 \right) $
Applying the \[a - {a_n} = - \left( {n - 1} \right)d\] formula in $ \left( 2 \right) $ we get,
 $ \Rightarrow \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{1}{{ - d}}\left( {\dfrac{{ - \left( {n - 1} \right)d}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}} \right) $
                                                                                             $ = \dfrac{{\left( {n - 1} \right)}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }} $
Hence, $ \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}} }} $

So, the correct answer is “Option B”.

Note: A common difference $ \left( d \right) $ is the difference between the first two terms.
Hence, the common difference $ d $ is given by \[{a_2} - {a_1}\; = {a_3} - {a_2} = \ldots . = {a_n} - {a_{n - 1}}\; = d\]
Multiplying by a minus sign, we get
 \[{a_1} - {a_2}\; = {a_2} - {a_3} = \ldots . = {a_{n - 1}} - {a_n}\; = - d\]
We know that \[{a_n} = a + \left( {n - 1} \right)d\]
 \[{a_n} = a + \left( {n - 1} \right)d \Rightarrow a - {a_n} = - \left( {n - 1} \right)d\]
These are the required formulae to obtain the answer. These are derived from the original formulae for our convenience.
Hence, $ \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}} }} $