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

Correct Answer: (C) I

The sequence consists of repeating alphabetical blocks of increasing length: block 1 is A (length 1), block 2 is AB (length 2), block 3 is ABC (length 3), and so forth. The total number of letters in the first $n$ blocks is given by $\frac{n(n+1)}{2}$$\frac{n(n+1)}{2}$. For $n = 13$, the cumulative count is $\frac{13\times 14}{2}=91$$\frac{13 \times 14}{2}$ = 91. The 14th block starts at position 92 with 'A'. To reach the 100th place, we need the $(100 - 91) = 9$$\text{th}$ letter of the 14th block, which is the 9th letter of the alphabet: 'I'.