document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1003729>Question 1003729</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The sum of the infinite geometric sequence  2 - 1/5 + 1/25 - 1/125 + .... is ______.\r<BR>\n" );
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document.write( "A. 14/5<BR>\n" );
document.write( "B. 6/5<BR>\n" );
document.write( "C. 6/13<BR>\n" );
document.write( "D. 11/6<BR>\n" );
document.write( "E. 9/4 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #620381 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=620381\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We have the first term, which is a = 2. We need the common ratio r.\r<BR>\n" );
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document.write( "Let's divide the second term over the first term<BR>\n" );
document.write( "r = (second term)/(first term)<BR>\n" );
document.write( "r = (-1/5)/(2)<BR>\n" );
document.write( "r = (-1/5)*(1/2)<BR>\n" );
document.write( "r = -1/10\r<BR>\n" );
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document.write( "Let's divide the third term over the second term<BR>\n" );
document.write( "r = (third term)/(second term)<BR>\n" );
document.write( "r = (1/25)/(-1/5)<BR>\n" );
document.write( "r = (1/25)*(-5/1)<BR>\n" );
document.write( "r = -5/25<BR>\n" );
document.write( "r = -1/5\r<BR>\n" );
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document.write( "Notice how r changes value. Since r is NOT the same each time, this means that this sequence is NOT a geometric sequence. So there has to be a typo somewhere. Please double check the problem. Thank you.\r<BR>\n" );
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document.write( "Edit: I realize that the sequence \" - 1/5 + 1/25 - 1/125 + ...\" is geometric (ignore the 2 up front). The first term of this sequence is a = -1/5 . The common ratio is...\r<BR>\n" );
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document.write( "r = (second term)/(first term)<BR>\n" );
document.write( "r = (1/25)/(-1/5)<BR>\n" );
document.write( "r = (1/25)*(-5/1)<BR>\n" );
document.write( "r = -5/25<BR>\n" );
document.write( "r = -1/5\r<BR>\n" );
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document.write( "Since |r| < 1, the infinite series converges to a fixed number, which is\r<BR>\n" );
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document.write( "S = a/(1-r)<BR>\n" );
document.write( "S = (-1/5)/(1-(-1/5))<BR>\n" );
document.write( "S = (-1/5)/(1+1/5)<BR>\n" );
document.write( "S = (-1/5)/(6/5)<BR>\n" );
document.write( "S = (-1/5)*(5/6)<BR>\n" );
document.write( "S = -1/6\r<BR>\n" );
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document.write( "So  2 - 1/5 + 1/25 - 1/125 + .... = 2 + (- 1/5 + 1/25 - 1/125 + ....) = 2 + (-1/6) = 11/6\r<BR>\n" );
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document.write( "Final answer: 11/6 </SPAN>\n" );
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