document.write( "Question 1076783: In an a.p the sum of the squares of the five consecutive series is 20time the square of the middle term,and the product of the five terms equals to 80.What is the middle term.
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Algebra.Com's Answer #691399 by ikleyn(52787)\"\" \"About 
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document.write( "Let \"a\" be the middle term of the AP, and \"d\" be the common difference.\r\n" );
document.write( "Then you have these equations\r\n" );
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document.write( "\"%28a-2d%29%5E2+%2B+%28a-d%29%5E2+%2B+a%5E2+%2B+%28a%2Bd%29%5E2+%2B+%28a%2B2d%29%5E2\" = \"20%2Aa%5E2\",    (1)    and\r\n" );
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document.write( "\"%28a-2d%29%2A%28a-d%29%2Aa%2A%28a%2Bd%29%2A%28a%2B2d%29\" = \"80\".             (2)\r\n" );
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document.write( "Simplify (1) and (2):\r\n" );
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document.write( "\"5a%5E2+%2B+10d%5E2\" = \"20%2Aa%5E2\"          (1')     and\r\n" );
document.write( "\"a%2A%28a%5E2-d%5E2%29%2A%28a%5E2-4d%5E2%29\" = \"80\"    (2')\r\n" );
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document.write( "Simplify (1'):\r\n" );
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document.write( "\"a%5E2+%2B+2d%5E2\" = \"4a%5E2\",   which is the same as  \"3a%5E2\" = \"2d%5E2\",  which is the same as\r\n" );
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document.write( "\"d%5E2\" = \"%283%2F2%29%2Aa%5E2\".               (1'')\r\n" );
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document.write( "Then (2') becomes (after substitution (1'') )\r\n" );
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document.write( "\"a%2A%28a%5E2+-+%283%2F2%29%2Aa%5E2%29%2A%28a%5E2+-+4%2A%283%2F2%29%2Aa%5E2%29\" = \"80\",   or\r\n" );
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document.write( "\"a%2A%28%28-1%2F2%29%2Aa%5E2%29%2A%28-5a%5E2%29\" = \"80\",   or\r\n" );
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document.write( "\"a%5E5\" = \"%2880%2A2%29%2F5\" = \"32\".\r\n" );
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document.write( "Then a = \"root%285%2C32%29\" = 2.\r\n" );
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document.write( "Answer.  The middle term under the question is 2.\r\n" );
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