If a1,a2,a3,…,a4001are term of an AP such  that 1a1a2+1a2a3+…+1a4000a4001=10a2 and  a2+a4000=50, then a1−a4001is equal to

Now,1a1a2+1a2a3+…+1a4000a4001=1da2−a1a1a2+a3−a2a2a3+…+a4001−a4000a4000a4001=1d1a1−1a2+1a2−1a3+…+1a4000−1a4001=1d1a1−1a4001=4000a1a4001=10       (given) ⇒ a1a4001=400                              ...(i)            a1+a4001=a2+a4000=50                .....(ii)∴ a1−a40012=a1+a40012−4a1a4001=(50)2−1600⇒ a1−a4001=30