document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=974847>Question 974847</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the sum of the second and third terms of a G.P is six times the fourth term. find the two possible values of the common difference. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #596712 by </B><A HREF=/tutors/aboutme.mpl?userid=amarjeeth123><B>amarjeeth123(569)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=amarjeeth123><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=596712\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the first term of the GP be a.<BR>\n" );
document.write( "Let the common ratio be r.<BR>\n" );
document.write( "second term=ar<BR>\n" );
document.write( "third term=ar^2<BR>\n" );
document.write( "fourth term=ar^3<BR>\n" );
document.write( "We have ar+ar^2=6ar^3<BR>\n" );
document.write( "1+r=6r^2<BR>\n" );
document.write( "6r^2-r-1=0<BR>\n" );
document.write( "6r^2-3r+2r-1=0<BR>\n" );
document.write( "3r(2r-1)+1(2r-1)=0<BR>\n" );
document.write( "(3r+1)(2r-1)=0<BR>\n" );
document.write( "r=-1/3 or r=1/2<BR>\n" );
document.write( "The two possible values of the common difference are r=-1/3 and r=1/2.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );