Question

The 288th term of the sequence a, b, b, c, c, c, d, d, d, d... is

• A. u
• B. v
• C. w
• D. x

Answer 1

The Correct Option is D

To solve the problem of determining the 288th term in the sequence, observe the pattern: a, b, b, c, c, c, d, d, d, d, ..., where each letter repeats the same number of times as its position in the alphabet (i.e., a = 1 time, b = 2 times, c = 3 times, etc.). We need to find the term number where each letter's repetition sum reaches 288.

Consider the cumulative sums:

• a: 1 (1st term)
• b: 2 + 1 = 3 (3rd term)
• c: 3 + 3 = 6 (6th term)
• d: 4 + 6 = 10 (10th term)...

Continue this series until reaching or exceeding 288. The sum of the first n natural numbers = n(n+1)/2.

We need to solve n(n+1)/2 = 288. Solve the quadratic equation:

n(n+1) = 576

n² + n - 576 = 0

Using the quadratic formula n = [-b ± √(b²-4ac)]/2a, with a=1, b=1, c=-576:

n = [-1 ± √(1 + 2304)]/2 = [-1 ± √2305]/2 ≈ 23.5

Since n must be an integer, try n = 23:

For n = 23, 23(23+1)/2 = 276

For n = 24, which starts at 277 and ends at:

n	cumulative sum
24	300

The terms 277 to 300 are 'x', thus the 288th term is 'x'.