jmc

Let $d$ be the common difference. Then $$\begin{array}{rl}\frac{1}{a_n{a}_{n+1}}& =\frac{1}{a_n(a_n+d)}\\ & =\frac{1}{d}\cdot \frac{d}{a_n(a_n+d)}\\ & =\frac{1}{d}\cdot \frac{(a_n+d)-a_n}{a_n(a_n+d)}\\ & =\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_n+d}\right)\\ & =\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right).\end{array}$$Thus, $$\begin{array}{rl}\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+\cdots +\frac{1}{a_{4000}{a}_{4001}}& =\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_2}\right)+\frac{1}{d}\left(\frac{1}{a_2}-\frac{1}{a_3}\right)+\cdots +\frac{1}{d}\left(\frac{1}{a_{4000}}-\frac{1}{a_{4001}}\right)\\ & =\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{4001}}\right)\\ & =\frac{1}{d}\cdot \frac{a_{4001}-a_1}{a_1{a}_{4001}}.\end{array}$$Since we have an arithmetic sequence, $a_{4001}-a_1=4000d,$ so $$\frac{1}{d}\cdot \frac{a_{4001}-a_1}{a_1{a}_{4001}}=\frac{4000}{a_1{a}_{4001}}=10.$$Hence, $a_1{a}_{4001}=\frac{4000}{10}=400.$

Then $$|a_1-a_{4001}{|}^2=a_1^2-2a_1{a}_{4001}+a_{4001}^2=(a_1+a_{4001}{)}^2-4a_1{a}_{4001}=50^2-4\cdot 400=900,$$so $|a_1-a_{4001}|=\boxed{30}.$