document.write( "Question 1185943: 1.the geometric mean between the first two terms is 32.if the third term is 4.find the first term \n" ); document.write( "

Algebra.Com's Answer #816852 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "the geometric mean between the first two terms $\"highlight%28of_a_geometric_progression%29\"$ is 32.
\n" ); document.write( "if the third term is 4, find the first term.
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\n" ); document.write( "\n" ); document.write( "            @MathLover1 incorrectly read, interpret and treat the problem.\r
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\n" ); document.write( "\n" ); document.write( "            Her solution and her answer are incorrect.\r
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\n" ); document.write( "\n" ); document.write( "            I came to bring a correct solution.\r
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document.write( "The three terms of the geometric progression are  a, ar, ar^2,  from the first term to the third, inclusive.\r\n" );
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document.write( "Since the problem talks about \"geometric mean\", it means that the first two terms are POSITIVE,\r\n" );
document.write( "and HENCE, all the terms of this GP are positive; in particular, \"a\" is positive and \"r\" is positive.\r\n" );
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document.write( "The geometric men of the first two terms is \r\n" );
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document.write( "    GM(a,ar) =  = .\r\n" );
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document.write( "So, the first equation is\r\n" );
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document.write( "     = 32.       (1)\r\n" );
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document.write( "The second equation is\r\n" );
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document.write( "     = 4.        (2)\r\n" );
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document.write( "Dividing equation (2) by equation (1), you get\r\n" );
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document.write( "    r^(3/2) =  =  = .\r\n" );
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document.write( "Hence,  r^(1/2) =   and  r = .\r\n" );
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document.write( "THEREFORE,  the first term  is   =  =  = 4*16 = 64.\r\n" );
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document.write( "ANSWER.  The first term of the geometric progression is 64.\r\n" );
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document.write( "CHECK.   Then the terms of the progression are  64,  = 16,   = 4,\r\n" );
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document.write( "         and the geometric mean of the first two terms is   =  =  =  = 32.    ! Correct !\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved (correctly).\r
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