Given the sequence A, B, C, C, D, D, C, C, C, D, D, D, D, ... — that is, one A, two B's, three C's, four D's, five E's and so on — the 240th letter in the sequence will be: A. V B. U C. T D. W

GATE 2010 · General Aptitude · Sequence Series · medium

Answer: The 240th letter in the sequence is V (option A), which is the 22nd letter of the alphabet; its group spans positions 232 to 253.

1. Express cumulative count after k groups: The total number of letters in the first k groups = 1+2+3+...+k = k(k+1)/2 = T(k). Position 240 lies in group k where T(k-1) < 240 <= T(k).
2. Solve k(k+1)/2 >= 240 for smallest integer k: k(k+1)/2 = 240 => k^2 + k - 480 = 0. Discriminant = 1 + 1920 = 1921. sqrt(1921) approx 43.83. k = (-1 + 43.83)/2 approx 21.4. So smallest integer k = 22.
3. Verify T(21) and T(22): T(21) = 21 x 22 / 2 = 231. T(22) = 22 x 23 / 2 = 253. Since T(21) = 231 < 240 <= 253 = T(22), position 240 lies in group 22.
4. Identify the 22nd letter of the alphabet: Group 22 corresponds to the 22nd letter of the alphabet. Counting: A(1), B(2), C(3), D(4), E(5), F(6), G(7), H(8), I(9), J(10), K(11), L(12), M(13), N(14), O(15), P(16), Q(17), R(18), S(19), T(20), U(21), V(22). The 22nd letter is V.