document.write( "Question 1140593: Find three numbers in arithmetic progression whose sum and product are six and six respectively \n" ); document.write( "

Algebra.Com's Answer #761121 by ikleyn(52788) $\"\"$ $\"About$

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document.write( "First of all, the answer is OBVIOUS:  the three terms are 1, 2 and 3.\r\n" );
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document.write( "The reversed order 3, 2 and 1 works, too.\r\n" );
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document.write( "The solution is OBVIOUS, too.\r\n" );
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document.write( "First step is to notice that the middle term (the 2-nd term) is one third of the sum, i.e. one third of 6, which is 2.\r\n" );
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document.write( "Then the first and the third terms are  (2+d)  and  (2-d), where \"d\" is the common difference.\r\n" );
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document.write( "From it, you have for the product of the three terms\r\n" );
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document.write( "    (2-d)*2*(2+d) = 6,   which implies\r\n" );
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document.write( "     4 - d^2 = 6/2 = 3,\r\n" );
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document.write( "     d^2 = 4-3 = 1,   hence\r\n" );
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document.write( "     d = +/-  = +/- 1,\r\n" );
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document.write( "which gives the answer.\r\n" );
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