Correct Option (C)

The given series AABABCABCDABCDE... is constructed by concatenating groups of letters. Each group consists of consecutive letters of the English alphabet, starting from 'A', with the length of the group increasing sequentially.

Group 1: A (1 letter)
Group 2: AB (2 letters)
Group 3: ABC (3 letters)
...
Group n: Consists of n letters (A to the nth letter of the alphabet)

The total number of letters up to the end of the nth group is the sum of the first n natural numbers, given by the formula $S_{n} = \frac{n (n + 1)}{2}$ .

To determine the group containing the 100th letter, we calculate cumulative sums:

For $n = 13$, $S_{13} = \frac{13(13 + 1)}{2} = \frac{13 \times 14}{2} = 91$. This indicates that the 13th group concludes at the 91st position in the series.
For $n = 14$, $S_{14} = \frac{14(14 + 1)}{2} = \frac{14 \times 15}{2} = 105$. This indicates that the 14th group concludes at the 105th position.

Since 91 < 100 < 105, the 100th letter is located within the 14th group of the series.

To ascertain its specific position within the 14th group, we subtract the total number of letters preceding this group from 100:

Position within 14th group = $100 - S_{13} = 100 - 91 = 9$.

The 9th letter of the English alphabet is I.

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

Incorrect Options:

Options (A) G, (B) H, and (D) J are incorrect. The calculation establishes that the 100th letter corresponds to the 9th position within the 14th group of the series. G is the 7th letter of the alphabet, H is the 8th letter, and J is the 10th letter. None of these align with the determined 9th position.