document.write( "Question 1144114: if k+1,2k-1, 3k+1,are three consecutive terms of a geometric progression, find the possible values of the common ratio?
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Algebra.Com's Answer #765108 by ikleyn(52790) $\"\"$ $\"About$

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document.write( "If  k+1, 2k-1, 3k+1  are three consecutive terms of a geometric progression, then\r\n" );
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document.write( "    the ratio    is equal to the ratio  ,\r\n" );
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document.write( "by the definition of a geometric progression.\r\n" );
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document.write( "Hence,\r\n" );
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document.write( "     = .\r\n" );
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document.write( "It implies\r\n" );
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document.write( "   (3k+1)*(k+1) = (2k-1)^2\r\n" );
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document.write( "   3k^2 + k + 3k + 1 = 4k^2 - 4k + 1\r\n" );
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document.write( "   k^2   - 8k = 0\r\n" );
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document.write( "   k*(k-8) = 0\r\n" );
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document.write( "which has two roots  k= 0  and  k= 8.\r\n" );
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document.write( "If k= 0, then the first and the second terms of the GP are  k+1 = 1  and  2k-1 = -1, so the common ratio is   = -1.\r\n" );
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document.write( "If k= 8, then the first and the second terms of the GP are  k+1 = 9  and  2k-1 = 15, so the common ratio is   = .\r\n" );
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document.write( "ANSWER.  Under given conditions, the common ratio may have one of the two values  -1  and/or  .\r\n" );
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