Algebra.Com's Answer #780936 by ikleyn(52790)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
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document.write( " This problem is nice. But it becomes nice only in hands of those who knows how to solve it and
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document.write( " knows how to present the solution to others in a way they see its beaty.\r
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document.write( " It is not the way @josgarithmetic tries to do it: he simply does not know the right way.\r
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document.write( " By following his way, you will NEVER feel its beauty.\r
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document.write( "Let \"a\" be the middle term of the progression, and let \"d\" be its common difference.\r\n" );
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document.write( "Then the first term of the AP is  = d-a, and its third term is  = d+a.\r\n" );
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document.write( "Since the sum ot the three terms is equal to 42, you have this equation\r\n" );
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document.write( " (a-d) + a + (a+d) = 42, \r\n" );
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document.write( "or 3a = 42, which immediately implies a = 42/3 = 14.\r\n" );
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document.write( "Thus we just found the middle terms, very quickly, easy and practically MENTALLY.\r\n" );
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document.write( "Next, regarding the product of the terms, we have this equation \r\n" );
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document.write( " (a-d)*(a+d) = 52,\r\n" );
document.write( "or\r\n" );
document.write( " a^2 - d^2 = 52.\r\n" );
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document.write( "Substitute here a = 14, the value found couple of lines above, and you will get\r\n" );
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document.write( " d^2 = a^2 - 52 = 14^2 - 52 = 144;\r\n" );
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document.write( "therefore\r\n" );
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document.write( " d = +/-  = +/- 12.\r\n" );
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document.write( "At this stage, the problem is just solved.\r\n" );
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document.write( "ANSWER. For the three terms, there are two possibilities.\r\n" );
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document.write( " 1) The progression is 14-12 = 2, 14 and 14+12 = 26 ( with d = 12 ), or\r\n" );
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document.write( " 2) The progression is 14+12 = 26, 14 and 14-12 = 2 ( with d = -12 ).\r\n" );
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document.write( " So, the second progression is simply the reversed first progression.\r\n" );
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document.write( "Solved.\r
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document.write( "It is how this problem is designed, is intended and is expected to be solved.\r
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