document.write( "Question 1121709: 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 \n" ); document.write( "

Algebra.Com's Answer #737704 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "There is not enough information to get a single answer. There are several arithmetic series that satisfy the given conditions.

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "That leaves several possible solutions with the common difference and first term both being whole numbers.

\n" ); document.write( "(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
\n" ); document.write( "90, 92, 94, ..., 100, 102, ..., 112.

\n" ); document.write( "(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
\n" ); document.write( "79, 83, 87, ..., 99, 103, ..., 123.

\n" ); document.write( "And so on. The largest possible whole number common difference, if the first term is also a whole number, is 18; the series is
\n" ); document.write( "2, 20, 38, ..., 92, 110, ..., 200.

\n" ); document.write( "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
\n" ); document.write( "101, 101, 101, ..., 101, 101, ..., 101.

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