document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=338482>Question 338482</A>: <I><SPAN STYLE=\"word-break: break-all;\"> write an equation for the following sequence:\r<BR>\n" );
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document.write( "0, 10, 15, 17.5 18.75<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 #242618 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=242618\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A bit tricky to explain this one, but recognize that you are adding 10, then half of that, 5, then half of that, 2.5, etc.\r<BR>\n" );
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document.write( "Whenever you add half as much, there's usually a (1/2)^n factor involved somewhere...here is a problem like that...\r<BR>\n" );
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document.write( "I get a-sub-n = 20 - 20(1/2)^(n-1) or 20(1 - 1/[2^(n-1)])...see if that works for you... </SPAN>\n" );
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