document.write( "Question 1202668: Which term of the series 2187,729,243....is 1/9 \n" ); document.write( "

Algebra.Com's Answer #837649 by Edwin McCravy(20056) $\"\"$ $\"About$

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document.write( "One way to solve it is to notice that to get from 2187 to 729 you have to divide\r\n" );
document.write( "2187 by 3.  And to get 729 to 243 you must also divide by 3.  So you can get the\r\n" );
document.write( "answer by asking this question:\r\n" );
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document.write( "If I start with 2187, how many times must I divide by 3 until I get 1/9?\r\n" );
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document.write( "Then the answer will be one more than that because the first time I divide\r\n" );
document.write( "will be to get the second term. \r\n" );
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document.write( "Cross multiplying\r\n" );
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document.write( "So it's 1 more than 9, so it's the 10th term.\r\n" );
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document.write( "That's not how you're supposed to do it, though.\r\n" );
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document.write( "You're supposed to use geometric series formula, like that baby in the\r\n" );
document.write( "picture up there did it.\r\n" );
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document.write( "with , , \r\n" );
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document.write( "and solve for n.  You can use logs to solve it if you like.\r\n" );
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document.write( "Edwin

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