document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170453>Question 1170453</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If S1=0.7 and S2=2.1, in a geometric series, determine the sum of the first 12 terms In the series  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #795328 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=795328\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> s=a(1-r^n)/(1-r)<BR>\n" );
document.write( "r=3 n=12<BR>\n" );
document.write( "s=0.7(1-3^12)/-2<BR>\n" );
document.write( "=0.7(265720)=186,004<BR>\n" );
document.write( "0.7<BR>\n" );
document.write( "2.1<BR>\n" );
document.write( "6.3<BR>\n" );
document.write( "18.9<BR>\n" );
document.write( "56.7<BR>\n" );
document.write( "170.1<BR>\n" );
document.write( "510.3<BR>\n" );
document.write( "1530.9<BR>\n" );
document.write( "4592.7<BR>\n" );
document.write( "13778.1<BR>\n" );
document.write( "41334.3<BR>\n" );
document.write( "124002.9<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );