document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=204901>Question 204901</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am not understanding sequences and series?  Can someone help me understand this better and show the steps.  Solve question given that a3  = 3 then a6= -81   Find r?  Find a1.  Thank you\r<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 #154643 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=154643\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> given that a3 = 3 then a6= -81 Find r? Find a1<BR>\n" );
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document.write( "Since you are looking for \"r\" your series is a geometric series.<BR>\n" );
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document.write( "a3 = a1*r^2 = 3<BR>\n" );
document.write( "a6 = a1*r^5 = -81<BR>\n" );
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document.write( "Divide a6 by a3 to get:<BR>\n" );
document.write( "r^3 = -27<BR>\n" );
document.write( "r = -3<BR>\n" );
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document.write( "Substitute r = -3 into a3 to solve for a1<BR>\n" );
document.write( "a1*(-3)^2 = 3<BR>\n" );
document.write( "a1 = 3/9 = 1/3<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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