Stuttering The Alphabet

There is 1 A, 2 Bs, 3 Cs, and so on.

If we got all of the way through the alphabet with this pattern, there would be $1+2+3+\cdots + 26 = \frac{26\times 27}{2} = 351$ letters. We are 63 short, so we take out the 26 Zs and 25 Ys, and are left with 300. Quite clearly, then, the 288th letter is one of the 24 Xs.

N(N+1)/2 = 288, solving this we get the value N is approximately equal to 24 and its not less than 24, so the 288th term is X

It was quite easy by considering and assigning the various alphabets a,b,c,.....,z as 1,2,3,.........,26 and then doing according to what is asked in the question.

Let a = 1, b =2, c =3, etc.

The 288th term (n) can be found using the arithmetic sequence : 1 + 2 + 3 + ... + n-1 + n = 288

n/2 (a + (n-1)d) = 288

n/2 (1 + (n-1)1) = 288

n (1 + n - 1) = 576

n(n) = 576

n^2 = 576

n = sqrt.(576)

n = 24

Therefore the 288th term will be the 24th letter in the alphabet, which is X.

The solution is X :)