SOLUTION: The sum of 12 terms of an arithmetic series is 1212. The first term “a” and the common difference “d” are both whole numbers. Write down the first three terms of the series

The formula for the sum of the first n terms of an arithmetic progression is


     = .


So, in your case it should be


     = 1212,   or    = 2424.


Notice that 2424 = 24*101, and 101 is a prime number.


It means that "n" must divide 101 or must divide some multiple of 101.



Now the first example of such arithmetic progression is THIS:

    101 terms, each equal to 12.

    It is arithmetic progression consisting of 101 equal terms;  each term is equal to 12; the common difference is 0 (zero).



The second example is 

    202 terms, each equal to 6.

    It is arithmetic progression consisting of 202 equal terms;  each term is equal to 6; the common difference is 0 (zero).



Having these two examples, you can easily construct other similar examples with the common difference of 0.



I will help you to construct one more example with the non-zero common difference.

    51-th term is 12;

    50-th term is 11;  52-th term is 13;

    49-th term is 10;  53-th term is 14.

    . . .   and  so  on  . . . . . . . . 


    1-st term is 12-50 = - 38;  101-th term is 12+50 = 62.


Notice that each pair of the terms in each line has half the sum of 12 (!)


Thus, the progression starts from -38; has the common difference of 1; the last, 101-th term is 62; the sum of the first 101 terms is


     =  =  =  = 12*101 = 1212.