Consider the following series:
A B C D .... X Y Z | Y X ...... B A | B C D ...... Y Z | Y X ..... C B A | B C ........ Y Z ....
Which letter occupies the 1000th positing in the above series ?

We have 3 patterns:
I: A B C D ... X Y Z, which occurs only once.
Y X ... B A, which repeats alternately.
B C ... Y Z, which repeats alternately.
Now, I: has 26 terms.
So, number of terms before the desired term = (999 - 26) = 973.
Each of the patterns which occurs after I: has 25 letters.
Now, 973 ÷ 25 gives quotient = 38 and remainder = 23.
Thus, the 1000th trem of the given series is the 24th term of the 39th pattern after I:.
Clearly, the 39th pattern is II: and its 24th term is B.