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

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, $\dfrac{1}{{\sqrt {{a_1}} + \sqrt {{a_2}} }} + \dfrac{1}{{\sqrt {{a_2}} + \sqrt {{a_3}} }} + ..... + \dfrac{1}{{\sqrt {{a_{n - 1}}} + \sqrt {{a_n}} }} = \dfrac{{n - 1}}{{\sqrt {{a_1}} + \sqrt {{a_n}} }}$