The sequence 2345, 23455, 234555, 2345555, ....... is formed by writing '234'…

Concept: When a sequence is built from terms whose digit-lengths increase by a constant amount (here +1 digit per term), the lengths themselves form an arithmetic progression (AP). To find which term contains a given overall digit position, use the AP sum formula S(n) = (n/2)[2a + (n-1)d] to get the cumulative digit count after n terms, then locate the position within that term.

Application:

1. Identify the pattern: each term is '234' followed by an increasing run of 5s — term 1 is 2345 (4 digits), term 2 is 23455 (5 digits), term 3 is 234555 (6 digits), and so on. So the term lengths 4, 5, 6, ... form an AP with first term a = 4 and common difference d = 1.

2. Write the cumulative digit count after n terms using the AP sum formula: S(n) = (n/2)[2(4) + (n-1)(1)] = n(n+7)/2.

3. Check n = 16: S(16) = 16(23)/2 = 184 digits — fewer than 200, so digit 200 is not yet reached.

4. Check n = 17: S(17) = 17(24)/2 = 204 digits — at least 200, so the 200th digit falls inside the 17th term.

5. The 17th term has length 3 + 17 = 20 digits: '234' followed by seventeen 5s. It occupies overall positions 185 to 204 (since the first 16 terms use up positions 1 to 184).

6. Within the 17th term, positions 185, 186, 187 hold '2', '3', '4', and positions 188 through 204 hold the seventeen 5s. Since 199 and 200 both lie in the range 188-204, both digits are 5.

Cross-check: Position 200 is 200 - 184 = 16 digits into the 17th term, i.e., the 16th digit of that term. The term's digits are 2, 3, 4 followed by seventeen 5s, so its 16th digit is the (16-3) = 13th of the seventeen 5s — still well inside the block of 5s, confirming both the 199th and 200th digits are 5.

So the last two digits (the 199th and 200th) of the sequence are 55.