Question 1099271
<br>The other tutor clearly misunderstood the problem.  If the sequence of digits repeated, the number would not be irrational....<br>
The given decimal is
0.54321 0 12345 00 54321 000 12345 0000 54321 00000 12345 000000 ...<br>
I have added spaces to make it easier to analyze the pattern of the digits.  If we take each "54321" as the beginning of a new sub-sequence, then...
the first sub-sequence has 13 terms - 2 sequences of 5 nonzero digits, and 1+2=3 zeros
the second sub-sequence has 17 terms - 2 sequences of 5 nonzero digits, and 3+4=7 zeros
the third sub-sequence has 21 terms - 2 sequences of 5 nonzero digits, and 5+6=11 zeros<br>
Clearly the numbers of digits in the successive sub-sequences form the arithmetic sequence
13, 17, 21, 25, ...<br>
To solve the problem, we want to know where the end of one of these sub-sequences is shortly before the 550th term of the given sequence.<br>
The n-th term of the arithmetic sequence 13, 17, 21, 25, ... is {{{13+4(n-1) = 13+4n-4 = 4n+9}}}.<br>
The sum of the first n terms of an arithmetic sequence is the average of the first term and last term, multiplied by the number of terms,  For this sequence, that is
{{{((13+4n+9)/2)*n = 2n^2+11n}}}<br>
By one method or another, we find that the largest n for which this is less than 550 is n=14, which gives us a sum of {{{2(14^2)+11(14) = 546}}}<br>
That means the 546th term is the last 0 in one of the sub-sequences.  The 550th term is 4 digits after that, which is the 4th digit in the "54321" that starts each sub-sequence.<br>
So the 550th digit in the decimal part of the given irrational number is "2".