Algebra.Com's Answer #102751 by Edwin McCravy(20060)![\"\"](\"/images/stars/stars-13.gif\")  
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Show that if the sum of an arithmetic series with an odd number of terms is 0, then one of the terms of the series must be 0. (Hint: let the number of terms be 2k + 1. Show that a1 = -kd. Then find the (k + 1)st term.)
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document.write( " Sn = 0 (given)\r\n" );
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document.write( " Sn =  [2a1 + (n-1)d] (formula for sum of the first n terms) \r\n" );
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document.write( " [2a1 + (n-1)d] = 0 (Both sides equal Sn) \r\n" );
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document.write( " 2a1 + (n-1)d = 0 (Divide both sides by \r\n" );
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document.write( " 2a1 + (2k+1-1)d = 0 (Substitute 2k+1 for n) \r\n" );
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document.write( " 2a1 + 2kd = 0 (Simplification)\r\n" );
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document.write( " 2a1 = -2kd (Subtract 2kd from both sides\r\n" );
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document.write( " a1 = -kd (Divide both sides by 2) \r\n" );
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document.write( "Now we find ak+1\r\n" );
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document.write( " an = a1 + (n-1)d (Formula for the nth term) \r\n" );
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document.write( " ak+1 = -kd + (k+1-1)d (Substituting k+1 for n \r\n" );
document.write( " and -kd for a1 \r\n" );
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document.write( " ak+1 = -kd + kd (Simplification) \r\n" );
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document.write( " ak+1 = 0 (Simplification) \r\n" );
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document.write( "Edwin \n" );
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