In the series AABABCABCDABCDE.., which letter appears at the 100th place?

The given series is: AABABCABCDABCDE…
The pattern being followed here is: A, AB, ABC, ABCD, ABCDE, …

Number of alphabets increase in arithmetic progression.

If we have an arithmetic series, 1, 2, 3 …, then:
Sum of first n terms = (n/2) (2a + (n-1)d), where ‘n’ is the number of terms in the series, ‘a’ the first term, and ‘d’ the difference between two terms i.e. 1.

So, n(n+1)/2 >= 100, solving/estimating we get n=14.

For n=13, we have sum = 13×14/2 = 91. so, 92nd is A, 93rd is B and so, 100th is I.