document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1096099>Question 1096099</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find x in the series 1+4+7+...+x=715 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #710561 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=710561\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Sum=(n/2)(a1+an); also an=a1+d(n-1)=1+3n-3=3n-2<BR>\n" );
document.write( "715=(n/2)(a1+an)<BR>\n" );
document.write( "1430=n(a1+an); a1=1<BR>\n" );
document.write( "1430=n*(an+1); substitute 3n-2 for an<BR>\n" );
document.write( "1430=n(3n-1)=3n^2-n<BR>\n" );
document.write( "3n^2-n-1430=0<BR>\n" );
document.write( "n=(1/6) 1 +/- sqrt(1+17160); sqrt(17161)=131<BR>\n" );
document.write( "n=+ (positive root only) (1/6)*132=22<BR>\n" );
document.write( "Therefore, 715=11(1+an)<BR>\n" );
document.write( "65=a1+an=1+an<BR>\n" );
document.write( "an=x=64  ANSWER<BR>\n" );
document.write( "The sum of 1+4+7+10+13+16+19+22+25+28+31+34+37+40+43+46+49+52+55+58+61+64=715 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );