Question 1121709
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There is not enough information to get a single answer.  There are several arithmetic series that satisfy the given conditions.<br>
The sum of the 12 terms is 1212, so the average is 101.  We can think of it as 6 pairs of numbers each with a sum of 202.<br>
So we know the sum of the 1st and 12th numbers is 202; and also that the sum of the 6th and 7th numbers is 202.<br>
That leaves several possible solutions with the common difference and first term both being whole numbers.<br>
(1) The two middle numbers could be 100 and 102; that makes the common difference 2; the first term is 100 minus 5 times the common difference; the last term is 102 plus 5 times the common difference; the series is
90, 92, 94, ..., 100, 102, ..., 112.<br>
(2) The two middle numbers could be 99 and 103; that makes the common difference 4; the first term is 100 minus 5 times the common difference; the last term is 102 plus 5 times the common difference; the series is
79, 83, 87, ..., 99, 103, ..., 123.<br>
And so on.  The largest possible whole number common difference, if the first term is also a whole number, is 18; the series is
2, 20, 38, ..., 92, 110, ..., 200.<br>
Since the problem says the common difference is a whole number (instead of a positive integer), you could even have the common difference 0; the series would be
101, 101, 101, ..., 101, 101, ..., 101.<br>