What is in the 200th position of 1234 12344 123444 1234444....?
3
4
5
7
Explanation:
The given series is 1234, 12344, 123444, 1234444, .....
So the number of digits in each term are 4, 5, 6, ... or (3 + 1), (3 + 2), (3 + 3), .....upto n terms =$3n+\frac{n(n+1)}{2}$
So $3n+\frac{n(n+1)}{2}$ <=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) So last digit is 4 and last two digits are 44.
The given series is 1234, 12344, 123444, 1234444, .....
So the number of digits in each term are 4, 5, 6, ... or (3 + 1), (3 + 2), (3 + 3), .....upto n terms =$3n+\frac{n(n+1)}{2}$
So $3n+\frac{n(n+1)}{2}$ <=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) So last digit is 4 and last two digits are 44.