$\text{ If }{\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,\dots .{\mathrm{a}}_{4001}\text{ are terms of an AP such that }\frac{1}{{\mathrm{a}}_1{\mathrm{a}}_2}+\frac{1}{{\mathrm{a}}_2{\mathrm{a}}_3}+\dots +\frac{1}{{\mathrm{a}}_{4000}{\mathrm{a}}_{4001}}=10\text{ and }{\mathrm{a}}_1+{\mathrm{a}}_{4001}=50\text{ then }\left|{\mathrm{a}}_1-{\mathrm{a}}_{4001}\right|\text{ is equal to }$

$\begin{array}{l}\frac{1}{d}\left[\frac{a_{4001}-a_1}{a_1{a}_{4001}}\right]=10\\ \Rightarrow \frac{4000}{{\mathrm{a}}_1{\mathrm{a}}_{4001}}=10 \left(\because a_{4001}=a_1+4000d\Rightarrow \frac{1}{d}\left(a_{4001}-a_1\right)=4000\right)\\ \Rightarrow {\mathrm{a}}_1{\mathrm{a}}_{4001}=400\\ \mathrm{Also} \mathrm{given} {\mathrm{a}}_1+{\mathrm{a}}_{4001}=50\\ \therefore {\left({\mathrm{a}}_1-{\mathrm{a}}_{4001}\right)}^2={\left({\mathrm{a}}_1+{\mathrm{a}}_{4001}\right)}^2-4{\mathrm{a}}_1{\mathrm{a}}_{4001}\\ \Rightarrow \left|{\mathrm{a}}_1-{\mathrm{a}}_{4001}\right|=30\end{array}$