Algebra.Com's Answer #833210 by ikleyn(52788)  You can put this solution on YOUR website! . \n" );
document.write( "In a geometric sequence of real numbers, the sum of the first \n" );
document.write( "two terms is 7 and the sum of the first six terms is 91. What is the sum of \n" );
document.write( "the first four terms? \n" );
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document.write( "We are given\r\n" );
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document.write( " + = 7 (1)\r\n" );
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document.write( " + + + + + = 91 (2) \r\n" );
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document.write( "In terms of \"a\" (the first term) and \"r\" (the common ratio) these equalities take the form\r\n" );
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document.write( " a + ar = 7, (3)\r\n" );
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document.write( " a + ar + + + + = 91 (4)\r\n" );
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document.write( "In (4), group the terms\r\n" );
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document.write( " (a + ar) + ( + ) + ( + ) = 91. (5)\r\n" );
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document.write( "Re-write (5) in an equivalent form\r\n" );
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document.write( " (a + ar) + + ) = 91. (6)\r\n" );
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document.write( "In (6), replace (a+ar) by the value of 7, based on (3). You will get\r\n" );
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document.write( " 7 + + = 91. (7)\r\n" );
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document.write( "In (7), divide both sides by 7\r\n" );
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document.write( " 1 + + = 13,\r\n" );
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document.write( "or\r\n" );
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document.write( " + - 12 = 0. (8)\r\n" );
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document.write( "This biquadratic equation (8) is the quadratic equation relative . Solve it by factoring\r\n" );
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document.write( " = 0\r\n" );
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document.write( "Since we are given that the progression is in real numbers, can not be zero \r\n" );
document.write( "(it is positive at any real r), we conclude that only possible value of is\r\n" );
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document.write( " = 3.\r\n" );
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document.write( "Then for the sum of the first four terms of this geometric progression we have\r\n" );
document.write( " = a + ar + + = (a + ar) + ( + ) = (a+ar) + .\r\n" );
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document.write( "We substitute here a+ar = 7 and = 3, and we get\r\n" );
document.write( " = 7 + 3*7 = 7 + 21 = 28.\r\n" );
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document.write( "ANSWER. The sum of the first 4 terms of this GP is 28.\r\n" );
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document.write( "Solved.\r \n" );
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