document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1041852>Question 1041852</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the sum of the geometric series for which a1 = 125, r = &#8534;, and n<BR>\n" );
document.write( "= 4.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #656783 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=656783\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It is a1[(1-r^n)/(1-r)]<BR>\n" );
document.write( "125*(1-(2/5)^4)/3/5;<BR>\n" );
document.write( "first deal with 1-(2/5)^4=625/625-16/625=609/625<BR>\n" );
document.write( "The numerator is 125*609/625, with 125 being 1/5 of 625, so the numerator is 609/5<BR>\n" );
document.write( "That is divided by3/5, so invert and multiply: 609/5*5/3=203<BR>\n" );
document.write( "The terms are:<BR>\n" );
document.write( "125<BR>\n" );
document.write( "50<BR>\n" );
document.write( "20<BR>\n" );
document.write( "8<BR>\n" );
document.write( "They add to 203. Check </SPAN>\n" );
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