A math question testing your series skill :)

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The position of the last occurrence of each letter is a triangular number. If the nth triangular number is x, it is given by x = n(n+1)/2. And so n in terms of x is given by the equation

n = -0.5 + 0.5*sqrt(1+8x)

Plugging x = 288 into the equation yields n = 23.5. This means the 288th term is between the 23rd and 24th triangular numbers. The 23rd letter is W and the 24th letter is X, so the 288th term is X.

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You are very correct :)

I haven't checked this yet, but I believe it is X.

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After understanding the question, I can only conclude that X marks the spot...

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The right answer will be W.

Each alphabet here is repeated the number of times of its positional sequence in the series of alphabets. 1 A , 2 Bs, 3 Cs and so on. So we want to know where 288 comes as a result in the sequence 1+2+3+.......... and so on.

This sequence goes in an arithmetic progression with a sum characteristics with a formula x=n*(n+1)/2 or 288=n*(n+1)/2

solving for n we get something that starts with 23 and ends with 24. W is a positional alphabet for 23 that will be start at a position 277 and will be repeated 23 times before the start of X that will start at a position of 301. So, the 288 th position will obviously be W and not X.

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If you will solve in detail... you will come to find out it is X.... It is a CAT question and the answer was X.... try solving once more.... I did solve first with the natural progression method.,.. and then I sat to check it down... and I got X

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