So the sequence k=1,2,3,4,5,...... and so on.
Sequence k is an arithmetic progression.
Since the number k appears k times, the sum of the sequence up to the kth term is 1000.
then by using the sum of an arithmetic progression, i used it to find n
T3=2, and 1+2=3
T6=3, and 1+2+3=6
T10=4, and 1+2+3+4=10
Hence you want T1000=1+2+3+4+....+n.
However, you do not know that T1000 is the last of a repeated list of the same number.
Following CaptainBlack's advice,
generated from the sum of terms, so it's the last 44.
Hence T991 is 45 and how many of those follow T990 ?