Question

If \[{a_1},\,\,{a_2},\,\,{a_3},\,....\,{a_n}\] are in A.P. where ${a_i} > 0$ for all $i$ then $\dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }}$ equals to:

Answer 1

Hint: In this question first we will multiply and divide every term of the equation with the conjugate of the denominator. Now, after the simplification we will get the difference of two consecutive terms in an A.P which is the same in denominator of every term. Now, we will simplify the equation we got. After the simplification we will put the formula of difference of two consecutive terms of an A.P.

Formula Used: The formula for nth term of an AP is given as ${a_n} = {a_1} + \left( {n - 1} \right)d$ where ${a_n}$ is the nth term, ${a_1}$ is the first term, $n$ is number of terms and $d$ is the common difference.

Complete step by step solution: The given equation is $\dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }}$ , where \[{a_1},\,\,{a_2},\,\,{a_3},\,....\,{a_n}\] are in A.P. and ${a_i} > 0$ for all $i$ .
Now, we will multiply and divide each term of the above equation with the conjugate of its denominator.
$\Rightarrow \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + .... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }}$
$\Rightarrow \dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} \times \dfrac{{\sqrt {{a_2}} - \sqrt {{a_1}} }}{{\sqrt {{a_2}} - \sqrt {{a_1}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} \times \dfrac{{\sqrt {{a_3}} - \sqrt {{a_2}} }}{{\sqrt {{a_3}} - \sqrt {{a_2}} }} + .... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} \times \dfrac{{\sqrt {{a_n}} - \sqrt {{a_{n - 1}}} }}{{\sqrt {{a_n}} - \sqrt {{a_{n - 1}}} }}$
Now, with the help of conjugate we will be able to remove the square root from the denominator.
$ \Rightarrow \dfrac{{\sqrt {{a_2}} - \sqrt {{a_1}} }}{{{a_2} - {a_1}}} + \dfrac{{\sqrt {{a_3}} - \sqrt {{a_2}} }}{{{a_3} - {a_2}}} + .... + \dfrac{{\sqrt {{a_n}} - \sqrt {{a_{n - 1}}} }}{{{a_n} - {a_{n - 1}}}}$
We know that the difference between two consecutive terms of an A.P is $d = {a_n} - {a_{n - 1}}$
Therefore, the above equation can be written as:
$ \Rightarrow \dfrac{{\sqrt {{a_2}} - \sqrt {{a_1}} }}{d} + \dfrac{{\sqrt {{a_3}} - \sqrt {{a_2}} }}{d} + .... + \dfrac{{\sqrt {{a_n}} - \sqrt {{a_{n - 1}}} }}{d}$
Now, take $d$ common from the above equation
$ \Rightarrow \dfrac{1}{d}\left[ {\sqrt {{a_2}} - \sqrt {{a_1}} + \sqrt {{a_3}} - \sqrt {{a_2}} + .... + \sqrt {{a_n}} - \sqrt {{a_{n - 1}}} } \right]$
Now, simplify the above equation
$ \Rightarrow \dfrac{1}{d}\left[ {\sqrt {{a_n}} - \sqrt {{a_1}} } \right]\_\_\left( 1 \right)$
We know that the formula of the nth term of an A.P is given by ${a_n} = {a_1} + \left( {n - 1} \right)d$ .
Now, we can write the difference between two consecutive term of an A.P as \[d = \dfrac{{{a_n} - {a_1}}}{{\left( {n - 1} \right)}}\] . Put the value of $d$ in the equation $\left( 1 \right)$ .
$ \Rightarrow \dfrac{{\left( {n - 1} \right)}}{{{a_n} - {a_1}}}\left[ {\sqrt {{a_n}} - \sqrt {{a_1}} } \right]$
Now, ${a_n} - {a_1}$ can be written as \[\left( {\sqrt {{a_n}} - \sqrt {{a_1}} } \right)\left( {\sqrt {{a_n}} + \sqrt {{a_1}} } \right)\] in the above equation:
$ \Rightarrow \dfrac{{\left( {n - 1} \right)}}{{\left( {\sqrt {{a_n}} - \sqrt {{a_1}} } \right)\left( {\sqrt {{a_n}} + \sqrt {{a_1}} } \right)}}\left[ {\sqrt {{a_n}} - \sqrt {{a_1}} } \right]$
$\Rightarrow \dfrac{{\left( {n - 1} \right)}}{{\sqrt {{a_n}} + \sqrt {{a_1}} }}$

Note: In this question the most important thing is the multiplication and division with the conjugate of the denominator of each term of the equation to remove the square root from the denominator. The other important thing is the manipulation of the formula of the nth term of an A.P to get a formula for the difference of two consecutive terms of an A.P.