document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1080886>Question 1080886</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given the arithmetic progression ;3,8,13...Find the least number of terms whose sum is greater than 80..<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #694980 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=694980\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> One can just write them out<BR>\n" );
document.write( "3+8+13=24<BR>\n" );
document.write( "+18=42<BR>\n" );
document.write( "+23=65<BR>\n" );
document.write( "+28=93, so the 6th term will sum to more than 80.<BR>\n" );
document.write( "The formula <BR>\n" );
document.write( "S=(n/2)(2a+(n-1)d)<BR>\n" );
document.write( "80=(n/2)(6+5(n-1))=(n/2)(6+5n-5)<BR>\n" );
document.write( "160=n(5n+1)<BR>\n" );
document.write( "5n^2+5n-160=0<BR>\n" );
document.write( "n^2+n-32=0<BR>\n" );
document.write( "does not factor<BR>\n" );
document.write( "n=(1/2)(-1+/- sqrt (1+128)); sqrt 129=11.36<BR>\n" );
document.write( "n=10.36/2=5.18.  This must be rounded to the next highest number or 6.<BR>\n" );
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