## Wednesday, 19 October 2011

### 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 23rd alphabet ie., ‘w’ ends at 276th position and series of next alphabet ie., ‘x’ starts from 277 and ends at 300th position. That means 288thterm is nothing but ‘x’.