document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=865729>Question 865729</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the sum of infinite geometric series -16+12-9+... \r<BR>\n" );
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document.write( "Thank you!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #521852 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=521852\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The first term is -16. So a = -16\r<BR>\n" );
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document.write( "The common ratio is 12/(-16) = -3/4. You find this by dividing any term by it's previous term. It helps to keep the common ratio as a fraction. So r = -3/4\r<BR>\n" );
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document.write( "We know a = -16 and r = -3/4. Because |r| < 1 is true, we know that the infinite geometric series converges to some fixed number (and it doesn't diverge).\r<BR>\n" );
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document.write( "We find this sum using the formula\r<BR>\n" );
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document.write( "S = a/(1 - r)\r<BR>\n" );
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document.write( "S = (-16)/(1 - (-3/4)) ... plug in a = -16 and r = -3/4\r<BR>\n" );
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document.write( "S = (-16)/(1 + 3/4)\r<BR>\n" );
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document.write( "S = (-16)/(4/4 + 3/4)\r<BR>\n" );
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document.write( "S = (-16)/(7/4)\r<BR>\n" );
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document.write( "S = (-16)*(4/7)\r<BR>\n" );
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document.write( "S = <font color=\"red\">-64/7</font>\r<BR>\n" );
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document.write( "This means -16+12-9+... = <font color=\"red\">-64/7</font>\r<BR>\n" );
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document.write( "Note: -64/7 = -9.14285714285714 approximately, but it's always a better idea to stick to the exact answer. This is because it's easier to get the approximate answer from the exact (it's usually harder to go from approximate to exact).\r<BR>\n" );
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