document.write( "Question 190402: Write the next term in the sequence. Then write a rule for the nth term.
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Algebra.Com's Answer #142900 by Edwin McCravy(20054)\"\" \"About 
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Write the next term in the sequence.
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document.write( "If it weren't for the signs,\r\n" );
document.write( "we would have the sequence \r\n" );
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document.write( "5, 10, 15, 20, ...\r\n" );
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document.write( "So let's first get the nth term of that sequence.\r\n" );
document.write( "The difference between any term and the preceding\r\n" );
document.write( "term is 5. So the common difference d = 5.  The\r\n" );
document.write( "nth term of an arithmetic sequence is given by:\r\n" );
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document.write( "\"a%5Bn%5D+=+a%5B1%5D+%2B+%28n-1%29%2Ad\"\r\n" );
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document.write( "\"a%5B1%5D+=+5\" and \"d+=+5\"\r\n" );
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document.write( "Substituting,\r\n" );
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document.write( "\"a%5Bn%5D+=+5+%2B+%28n-1%29%2A5\" \r\n" );
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document.write( "\"a%5Bn%5D+=+5+%2B+5%28n-1%29\"\r\n" );
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document.write( "\"a%5Bn%5D+=+5+%2B+5n-5\"\r\n" );
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document.write( "\"a%5Bn%5D+=+5n\"\r\n" );
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document.write( "Now we have to tack on a factor that\r\n" );
document.write( "will cause the signs to alternate:\r\n" );
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document.write( "The alternating sign factors are \"%28-1%29%5En\"\r\n" );
document.write( "and \"%28-1%29%5E%28n%2B1%29\", depending on whether\r\n" );
document.write( "the first term is negative or positive. Since \r\n" );
document.write( "the first term is -5, a negative number, we\r\n" );
document.write( "use \"%28-1%29%5En\" for the alternating sign\r\n" );
document.write( "factor.  So if we use (capital \"A\") \"A%5Bn%5D\" \r\n" );
document.write( "for the nth term, we just multiply the nth term\r\n" );
document.write( "for \"a%5Bn%5D\" (small \"a\") above by \"%28-1%29%5En\" \r\n" );
document.write( "and get\r\n" );
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document.write( "\"A%5Bn%5D+=+%28-1%29%5En%2A%285n%29\"\r\n" );
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document.write( "Edwin
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