Series - 2 (SNAP-2008)

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 


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^thterm is nothing but ‘x’.

Answer (D)