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 (E) y

Number of repetitions of alphabet ‘a’ = 1

**Solution follows here:**
Solution:

The given series is a,b,b,c,c,c,d,d,d,d,...Number of repetitions of alphabet ‘a’ = 1

Number of repetitions of alphabet ‘b’
= 2

Number of repetitions of alphabet ‘c’
= 3, and so on and so forth..

The counts of alphabets follow a
sequence 1,2,3,…

Position of last ‘b’ is 3, which is
nothing but sum of ‘a’ counts and ‘b’ counts (ie., 1+2). And also observe that
alphabet number of ‘b’ is ‘2’.

Similarly, position of last ‘c’ =
1+2+3 = 6. Observe that alphabet number of ‘c’ is ‘3’

From the above, the conclusion is position
of last n’th alphabet is the sum of first ‘n’ integers ie., 1+2+3+…+n = n(n+1)/2.

We need the alphabet in position 288.

∴ n(n+1)/2 = 288 => n(n+1) = 576

Try out some multiplications, and we
get the following nearest values:

(23 * 24)/2 = 552/2 = 276 and (24 *
25)/2 = 600/2 = 300

288 is between 276 and 300, hence the
alphabet number ‘n’ is between 23 and 24.

∴Series of 23

^{rd}alphabet ie., ‘w’ ends at 276^{th}position and series of next alphabet ie., ‘x’ starts from 277 and ends at 300^{th}position. That means 288^{th}term is nothing but ‘x’.**Answer (D)**
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