In the series AABABCABCDABCDE..., which letter appears at the 100th place?
- (a) G
- (b) H
- (c) I ✓ UPSC's answer
- (d) J
Why the answer is (c)
• The series is formed by repeating blocks of increasing length: 'AA' (2 letters), 'AAB' (3 letters), 'AABC' (4 letters), and so on, where the n-th block has length n+1 and ends with the (n-1)-th letter of the alphabet.
• To find the 100th position, we sum the lengths of consecutive blocks until the cumulative total reaches or exceeds 100.
• The sum of the first k blocks is given by the formula S_k = k(k+3)/2, derived from summing an arithmetic progression starting at 2 with a common difference of 1.
• Solving k(k+3)/2 = 100 gives k^2 + 3k - 200 = 0, which yields k ≈ 12.3, indicating the 100th letter lies within the 13th block.
• The sum of the first 12 blocks is 12(15)/2 = 90, meaning the 13th block starts at position 91 and has a length of 14 (positions 91 to 104).
• The 13th block follows the pattern 'AABCDEFGHIJKLMN', so the 10th letter in this block (position 91 + 9 = 100) is 'I'.
Why the other options are wrong
- (a) G
- Option G is the 7th letter of the alphabet, which would appear at position 97 in the 13th block, not 100.
- (b) H
- Option H is the 8th letter of the alphabet, which would appear at position 98 in the 13th block, not 100.
- (d) J
- Option J is the 10th letter of the alphabet, which would appear at position 101 in the 13th block, not 100.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.