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**Hint**: In this Question firstly we will use \[\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ..... + \dfrac{1}{{{a_{n - 1}}{a_n}}} = \dfrac{{n - 1}}{{{a_1}{a_n}}}\] and equate the right hand side of equation with 10. After that we will substitute the value of n that is 4001 in this case. We will also use the formula \[{a_n} = a + \left( {n - 1} \right)d\] to calculate \[d\] then calculate the value of \[{a_1}\] and \[{a_{4001}}\] to find the result.

Formula used: Here we have used the formula \[{a_n} = a + \left( {n - 1} \right)d\] .

**:**

__Complete step-by-step answer__It is given that \[{a_1},{a_2},{a_3},....,{a_{4001}}\] are terms of an A.P such that \[\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ..... + \dfrac{1}{{{a_{4000}}{a_{4001}}}} = 10\] --equation2 and \[{a_2} + {a_{4000}} = 50,\] and we have to find the value of \[\left| {{a_1} - {a_{4001}}} \right|\] .

As we know that, \[\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ..... + \dfrac{1}{{{a_{n - 1}}{a_n}}} = \dfrac{{n - 1}}{{{a_1}{a_n}}}\] --equation1

By equating equation 1 and 2 we get,

\[ \Rightarrow \dfrac{{n - 1}}{{{a_1}{a_n}}} = 10\]

Here, \[n = 4001\]

So, by substituting the value of n we get,

\[ \Rightarrow \dfrac{{4001 - 1}}{{{a_1}{a_{4001}}}} = 10\]

After solving we get,

\[ \Rightarrow 400 = {a_1}{a_{4001}}\]

\[ \Rightarrow 400 = {a_1}\left( {{a_1} + 4000d} \right)\] --equation3

As, we are also given with \[{a_2} + {a_{4000}} = 50\]

\[ \Rightarrow {a_1} + d + {a_1} + 3999d = 50\]

\[ \Rightarrow 2{a_1} + 4000d = 50\]

\[ \Rightarrow {a_1} + 2000d = 25\]

\[ \Rightarrow {a_1} = 25 - 2000d\] --equation4

Now, we will substitute the value of \[{a_1}\] from equation4 to equation3 we get,

\[ \Rightarrow 400 = \left( {25 - 2000d} \right)\left( {25 - 2000d + 4000d} \right)\]

Simplifying further,

\[ \Rightarrow 400 = \left( {25 - 2000d} \right)\left( {25 + 2000d} \right)\]

As we know \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]

\[ \Rightarrow 400 = \left[ {{{\left( {25} \right)}^2} - {{\left( {2000d} \right)}^2}} \right] \]

By opening squares we get,

\[ \Rightarrow 400 = 625 - 4000000{d^2}\]

\[ \Rightarrow 4000000{d^2} = 625 - 400\]

\[ \Rightarrow 4000000{d^2} = 225\]

On simplifying we get,

\[ \Rightarrow {d^2} = \dfrac{{225}}{{4000000}}\]

Dividing both numerator and denominator by 25.

\[ \Rightarrow {d^2} = \dfrac{9}{{160000}}\]

Taking square root on both sides,

\[ \Rightarrow d = \pm \left( {\dfrac{3}{{400}}} \right)\]

Here, we will take \[d = \dfrac{3}{{400}}\]

Then we will calculate \[{a_1} = 25 - 2000d\] and we will substitute the value of d that is

\[ \Rightarrow {a_1} = 25 - 2000 \times \dfrac{3}{{400}}\]

On simplifying we get \[ \Rightarrow {a_1} = 10\]

And we will also calculate \[{a_{4001}} = {a_1} + 4000d\] substituting the values of \[{a_1}\] and d we get,

\[{a_{4001}} = 10 + 4000 \times \dfrac{3}{{400}}\]

On simplifying we get,

\[{a_{4001}} = 40\]

Now we will calculate \[\left| {{a_1} - {a_{4001}}} \right|\] by substituting the values we get,

\[ \Rightarrow \left| {{a_1} - {a_{4001}}} \right| = \left| {10 - 40} \right| = \left| { - 30} \right|\]

\[ \Rightarrow \left| {{a_1} - {a_{4001}}} \right| = 30\]

Here, we will take \[d = \dfrac{{ - 3}}{{400}}\]

Then we will calculate \[{a_1} = 25 - 2000d\] and we will substitute the value of d that is

\[ \Rightarrow {a_1} = 25 - 2000 \times \dfrac{{ - 3}}{{400}}\]

On simplifying we get \[ \Rightarrow {a_1} = 40\]

And we will also calculate \[{a_{4001}} = {a_1} + 4000d\] substituting the values of \[{a_1}\] and d we get,

\[{a_{4001}} = 40 + 4000 \times \dfrac{{ - 3}}{{400}}\]

On simplifying we get,

\[{a_{4001}} = 10\]

Now we will calculate \[\left| {{a_1} - {a_{4001}}} \right|\] by substituting the values we get,

\[ \Rightarrow \left| {{a_1} - {a_{4001}}} \right| = \left| {40 - 10} \right| = \left| {30} \right|\]

\[ \Rightarrow \left| {{a_1} - {a_{4001}}} \right| = 30\]

**So, the correct answer is “Option B”.**

**Note**: In these types of questions, we must remember the formulas of an A.P. series to make the equation simpler and easy to solve. Using the A.P. series formula we can find the values of variables in the series and then use them to calculate desired results asked in the problem.

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