document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187227>Question 1187227</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The first term of geometric progression is 300 and the fourth term is -8/5. Find the common ratio and the sum to infinity. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #818336 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=818336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the ratio is the cube root of (-1.6/300), since this is a geometric series.<BR>\n" );
document.write( "that is -0.1747<BR>\n" );
document.write( "sum=a/(1-r)<BR>\n" );
document.write( "=300/1.1747<BR>\n" );
document.write( "=255.38 sum at infinity.<BR>\n" );
document.write( "terms<BR>\n" );
document.write( "300<BR>\n" );
document.write( "52.41 minus (247.59)<BR>\n" );
document.write( "9.16 (256.75)<BR>\n" );
document.write( "1.60 minus (255.15)<BR>\n" );
document.write( "0.28 255.43<BR>\n" );
document.write( "0.05 minus (257.38) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );