If ${a_1},{a_2},...{a_n}$are in A.P, then $\dfrac{1}{{{a_1}{a_2}}},\dfrac{1}{{{a_2}{a_3}}},...~\dfrac{1}{{{a_{n - 1}}{a_n}}}$is
A. $\dfrac{{n - 1}}{{{a_1}{a_n}}}$
B. $\dfrac{1}{{{a_1}{a_n}}}$
C. $\dfrac{{n + 1}}{{{a_1}{a_n}}}$
D. $\dfrac{n}{{{a_1}{a_n}}}$

Hint: In this question, we will find the common difference of the given A.P then substitute all the values and simply after that we will use the formula of the $n^{th}$ term of A.P to find the result.

Formula Used:
1. ${a_n} = a + \left( {n - 1} \right)d$
2. d = Second term - First term

Complete step by step solution:
We are given that ${a_1},{\text{ }}{a_2}, \ldots {a_n}$ are in A.P
Let $d$ be the common difference between A.P
$d{\text{ }} = {\text{ }}{a_2} - {a_1}\; = {\text{ }}{a_3} - {a_2}\; = {\text{ }}{a_4} - {a_3}\; \ldots . = {\text{ }}{a_n} - {a_{n - 1}}$
So,
$
  \dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ...... + \dfrac{1}{{{a_{n - 1}}{a_n}}} = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{d}{{{a_1}{a_2}}} + \dfrac{d}{{{a_2}{a_3}}} + .... + \dfrac{d}{{{a_{n - 1}}{a_n}}}} \right) \\
   = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{{\left( {{a_2} - {a_1}} \right)}}{{{a_1}{a_2}}} + \dfrac{{\left( {{a_3} - {a_2}} \right)}}{{{a_2}{a_3}}} + ...\dfrac{{\left( {{a_n} - {a_{n - 1}}} \right)}}{{{a_{n - 1}}{a_n}}}} \right) \\
   = \left( {\dfrac{1}{d}} \right)\left( {\left( {\dfrac{{{a_2}}}{{{a_1}{a_2}}} - \dfrac{{{a_1}}}{{{a_1}{a_2}}}} \right) + \left( {\dfrac{{{a_3}}}{{{a_2}{a_3}}} - \dfrac{{{a_2}}}{{{a_2}{a_3}}}} \right) + ....\left( {\dfrac{{{a_n}}}{{{a_{n - 1}}{a_n}}} - \dfrac{{{a_{n - 1}}}}{{{a_{n - 1}}{a_n}}}} \right)} \right) \\
   = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{1}{{{a_1}}} - \dfrac{1}{{{a_2}}} + \dfrac{1}{{{a_2}}} - \dfrac{1}{{{a_3}}} + ... + \dfrac{1}{{{a_{n - 1}}}} - \dfrac{1}{{{a_n}}}} \right) \\
 $
Further solving we get

$
  \dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ...... + \dfrac{1}{{{a_{n - 1}}{a_n}}} = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{1}{{{a_1}}} - \dfrac{1}{{{a_n}}}} \right) \\
   = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{{{a_n} - {a_1}}}{{{a_1}{a_n}}}} \right) \\
   = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{{{a_1} + \left( {n - 1} \right)d - {a_1}}}{{{a_1}{a_n}}}} \right)~~~~~\left( {\because {a_n} = a + \left( {n - 1} \right)d} \right) \\
   = \left( {\dfrac{1}{d}} \right)\left( {\dfrac{{\left( {n - 1} \right)d}}{{{a_1}{a_n}}}} \right) \\
 $
Furthermore, solving we get
$\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ...... + \dfrac{1}{{{a_{n - 1}}{a_n}}} = \dfrac{{n - 1}}{{{a_1}{a_n}}}$

Option ‘A’ is correct

Note: To make the problem simpler and easier to solve, we must recall the formulas of an A.P. series in these types of situations. We may get the values of variables in the series using the A.P. series formula, and then use them to calculate the desired solutions to the problem. We come across quite a few examples of advancement in our daily lives. For example, class roll numbers.