Quantitative Aptitude :: Numbers - Discussion

the given series is 2345 23455 234555......

So the number of digits in each term are 4, 5, 6, ... or (3  1), (3  2), (3  3),......upto nth terms are=3n [n(n 1)/2]

so 3n [n(n 1)/2]>=200

For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184. And 16 them contains 16  3 = 19 digits.

Now 17 term contains 20 digits and 123444......4(17times)...  last two digits are 55.

Now 17 term contains 20 digits and 2345555......4(17times)...  last two digits are 55.

sorry its mistake in solution..:-)

Solution:  

1st group 2345 number of digit 4

2nd group 23455 number of digit 5

3rd group 234555 number of digit 6

so at the end of each group, there will be series of 5s.

So the answer is 55