# AABABCABCDABCDE... which letter occupies 100th position?

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### 2 Answers

We'll, it is easy! I've explained above that the letter D corresponds to the 10th position in this series of letter.

Since the sum 1 + 2 + 3 + ... + 13 + 14 yields 105, which is larger than 100, then the ultimate group of letters is:

(A)(AB)(ABC)(ABCD)(ABCDE)....(ABCDEFGHIJKLM)(ABCDEFGHI)

Therefore the letter M is located on the 91st position. From here, we'll start another sequence and we'll count A=92, B=93, C=94,D=95,E=96,F=97,G=98,H=99,I=100.

Logically, that's why the letter I is located on the 100th position.

We notice that we can create groups in the given series of letters:

(A)(AB)(ABC)(ABCD)(ABCDE)....

We notice that the first B letter occupies the 3rd position and the 1st C letter occupies the 6th position.

To determine the position of the first D letter:

1 + 2 + 3 + 4 = 10

The 100th position will be found using the formula that gives the sum of the first n natural numbers:

Sn = (1+n)*n/2

100 = (1 + x)*x/2

200 = x + x^2

x^2 + x - 200 = 0

x1 = [-1+sqrt(1+800)]/2

x1 = 13 approx.

1 + 2 + 3 + ... + 13 = 91

**We notice that the letter that occupies the 100th position will be the 9th letter of the alfabet:ABCDEFGHI, that means that it is the letter I.**