What is the 144th letter in this pattern?
A B B C C C D D D D E E E E E.....
To find the 144th letter in this pattern, we can observe that the number of repetition of each letter increases by one with each letter. So, the first letter "A" appears once, the second letter "B" appears twice, the third letter "C" appears three times, and so on.
To determine which letter appears 144 times, we need to find the smallest number "n" for which the sum of the first n natural numbers is greater than or equal to 144.
The sum of the first n natural numbers can be calculated using the formula: sum = n * (n + 1) / 2.
Let's find the value of n that satisfies the condition:
n * (n + 1) / 2 ≥ 144
Simplifying the inequality:
n^2 + n ≥ 288
n^2 + n - 288 ≥ 0
To solve this quadratic inequality, we can factorize if possible or use the quadratic formula. In this case, factoring is not possible, so we will use the quadratic formula:
n = (-1 ± sqrt(1^2 - 4 * 1 * (-288))) / 2
= (-1 ± sqrt(1 + 1152)) / 2
= (-1 ± sqrt(1153)) / 2
Since we need the positive value of n, we take the positive square root:
n = (-1 + sqrt(1153)) / 2
Evaluating this expression gives us approximately:
n ≈ 17.95
Since n represents the number of letters before the letter that appears 144 times, we can round it up to the next whole number:
n ≈ 18
Therefore, the letter that appears 144 times in the pattern is the 18th letter in the sequence.
In this case, the 18th letter is "E".