Do you know its Property-3

$As\quad it\quad is\quad increasing\quad AP\quad so\quad d\quad is\quad +ve\\ \\ \cfrac { 1 }{ { a }_{ r }{ a }_{ r+1 } } =\cfrac { 1 }{ { a }_{ r }({ a }_{ r }+d) } =\cfrac { 1 }{ d } \left( \cfrac { 1 }{ { a }_{ r } } -\cfrac { 1 }{ { a }_{ r }+d } \right) =\cfrac { 1 }{ d } \left( \cfrac { 1 }{ { a }_{ r } } -\cfrac { 1 }{ { a }_{ r+1 } } \right) \\ \\ \therefore \sum _{ r=1 }^{ 4000 }{ \cfrac { 1 }{ { a }_{ r }{ a }_{ r+1 } } = } \cfrac { 1 }{ d } \left( \cfrac { 1 }{ { a }_{ 1 } } -\cfrac { 1 }{ { a }_{ 4000+1 } } \right) =\cfrac { 1 }{ d } \left( \cfrac { 1 }{ { a }_{ 1 } } -\cfrac { 1 }{ { a }_{ 4001 } } \right) =10\\ \\ { a }_{ 1 }={ a }_{ 2001 }-2000d=25-2000d\\ \\ { a }_{ 4001 }={ a }_{ 2001 }+2000d=25+2000d\\ \\ \therefore 10=\cfrac { 1 }{ d } \left( \cfrac { 1 }{ { a }_{ 1 } } -\cfrac { 1 }{ { a }_{ 4001 } } \right) =\cfrac { 1 }{ d } \left( \cfrac { 1 }{ 25-2000d } -\cfrac { 1 }{ 25+2000d } \right) =\cfrac { 4000 }{ { (25) }^{ 2 }-{ (2000d) }^{ 2 } } \\ \\ \Rightarrow { \left( 2000d \right) }^{ 2 }=625-400=225\\ \\ \Rightarrow 2000d=15\left( as\quad d\quad is\quad +ve \right) \\ \\ \left| { a }_{ 1 }-{ a }_{ 4001 } \right| =\left| -4000d \right| =\left| -30 \right| =30$

There's another cute property of AP- $\huge\boxed {\frac{(k-1)}{a_1 a_k}+\frac{1}{a_k a_{k+1}}=\frac{k}{a_1 a_{k+1}}}$ $\textbf{(Try proving it)}$ Using this we can telescope the sum, $\Rightarrow \frac{4000}{a_1 a_{4001}}=10\Rightarrow a_1 a_{4001}=400$ Now $\large 2a_{2001}=50=a_1+a_{4001}\text {(why)}$ $\Rightarrow (a_1-a_{4001})^2=(a_1+a_{4001})^2-4 a_1 a_{4001}=900$ $\large |a_1-a_{4001}|=\boxed{30}$