document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=873553>Question 873553</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the sum of the infinite geometric series\r<BR>\n" );
document.write( "\n" );
document.write( "-12 + 48/5 - 192/25 + ...\r<BR>\n" );
document.write( "\n" );
document.write( "here are the answer choices\r<BR>\n" );
document.write( "\n" );
document.write( "A) negative 40/3<BR>\n" );
document.write( "B)negative 80/3<BR>\n" );
document.write( "C) negative 20/3<BR>\n" );
document.write( "D) negative 10/3\r<BR>\n" );
document.write( "\n" );
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 #526944 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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/cgi-bin/show-face.mpl?user=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=526944\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> S=-12*(1 - (-4/5)^20)/(1 - (-4/5))=-6  2/3\r<BR>\n" );
document.write( "\n" );
document.write( "S=t*(1 - r^n)/(1 - r)<BR>\n" );
document.write( "r=-4/5<BR>\n" );
document.write( "c) -20/3 is  -6 2/3 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );