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

The correct option is I.

To find the letter at the 100th place, we first need to understand the pattern of the given series:
A, AB, ABC, ABCD, ABCDE, ...

We can divide the series into groups:
- Group 1: A (1 letter)
- Group 2: AB (2 letters)
- Group 3: ABC (3 letters)
- Group 4: ABCD (4 letters)
- Group n: A, B, C, ... (n letters)

The total number of letters up to the end of the n-th group is given by the sum of the first n natural numbers. We can represent this sum using the formula:

$S_n=\frac{n(n+1)}{2}$

We want to find the group that ends just before or at the 100th letter. Let's test values of n:
If we choose n = 13:

$S_{13}=\frac{13(13+1)}{2}=\frac{13\left(14\right)}{2}=91$

This tells us that the 13th group ends at the 91st letter of the series. Since the 13th letter of the alphabet is M, the 91st position is occupied by the letter M.

The 14th group begins from the 92nd position, starting again with the letter A:
- 92nd letter: A
- 93rd letter: B
- 94th letter: C
- 95th letter: D
- 96th letter: E
- 97th letter: F
- 98th letter: G
- 99th letter: H
- 100th letter: I

Therefore, the letter appearing at the 100th place is I.