a, b, b, c, c, c, d, d, d, d, . . . . . . Find the 288th letter of this series.
Observe that each letter appeared once, twice, thrice .... They form an arithmetic progression. 1+2+3......
We know that sum of first n natural numbers = n(n+1)2n(n+1)2
So n(n+1)2n(n+1)2 ≤≤ 288
For n = 23, we get 276. So for n = 24, the given series crosses 288.
Ans is X.