document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=859437>Question 859437</A>: <I><SPAN STYLE=\"word-break: break-all;\"> please help to solve the following\r<BR>\n" );
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document.write( "the sum of the first and the third terms of a geometric progression is 40. the sum of the second and the fourth terms is 120.  Determine the first term and the common ratio of the progression. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #517751 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=517751\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the terms will be<BR>\n" );
document.write( "a,ar,ar^2,ar^3\r<BR>\n" );
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document.write( "Condition I\r<BR>\n" );
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document.write( "a+ar^2=40<BR>\n" );
document.write( "a*(1+r^2)=40<BR>\n" );
document.write( "(1+r^2) =40/a\r<BR>\n" );
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document.write( "condition II\r<BR>\n" );
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document.write( "ar+ar^3=120\r<BR>\n" );
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document.write( "ar(1+r^2)=120\r<BR>\n" );
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document.write( "ar(40/a) =120\r<BR>\n" );
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document.write( "40r =120\r<BR>\n" );
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document.write( "r=3\r<BR>\n" );
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document.write( "plug r to get a </SPAN>\n" );
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