document.write( "Question 1012889: Find five consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156. \n" ); document.write( "

Algebra.Com's Answer #629010 by ikleyn(52787) $\"\"$ $\"About$

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\n" ); document.write( "Find five consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156.
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document.write( "You can write these terms in this way\r\n" );
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document.write( "a-2d, a-d a, a+d, a+2d\r\n" );
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document.write( "where a if the third (the middle of 5) term, and d is the common difference.\r\n" );
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document.write( "Then the sum of 5 terms is 5a, and you easily find a =  = 23.\r\n" );
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document.write( "To find d, use the second part of the condition\r\n" );
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document.write( " = 1156,   or\r\n" );
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document.write( " =  = 578,   or\r\n" );
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document.write( " = 578 -  = 49.\r\n" );
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document.write( "Hence, d = +/-7. \r\n" );
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document.write( "Although you have two solutions for d, the progression is actually unique.\r\n" );
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document.write( "It is 9, 16, 23, 30, 37.\r\n" );
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