document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=856796>Question 856796</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the third term of a GP is 36 and the 6th term is 243/2 find the first term,the common ratio and the sum of the 8th terms </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #516222 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=516222\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Tn = t1 * r^(n - 1)<BR>\n" );
document.write( "T3=36= t1 * r^(2)<BR>\n" );
document.write( "T6 =243/2= t1 * r^(5)<BR>\n" );
document.write( "243/2= t1 * r^(5),<BR>\n" );
document.write( "36= t * r^(2)<BR>\n" );
document.write( "r = 3/2,   t1 = 16<BR>\n" );
document.write( "S=t1*(1 - r^n)/(1 - r)<BR>\n" );
document.write( "S=16*(1 - (3/2)^8)/(1 - 3/2)<BR>\n" );
document.write( "S=6305/8<BR>\n" );
document.write( "S=788 1/8 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );