What is in the 200th position of 1234 12344 123444 1234444....?

What is in the 200th position of 1234 12344 123444 1234444....?

Answer: 4











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+n(n+1)2$3n+{\displaystyle \frac{n(n+1)}{2}}$
So 3n+n(n+1)2≤200$3n+{\displaystyle \frac{n(n+1)}{2}}\le 200$
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$123\underset{17\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{s}}{\underbrace{\mathrm{444......4}}}$.  So last digit is 4 and last two digits are 44.