document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=855042>Question 855042</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is Sn of the geometric series with a1=3, an=-38, and n=8?\r<BR>\n" );
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document.write( "A) -178<BR>\n" );
document.write( "B)-102<BR>\n" );
document.write( "C)-140<BR>\n" );
document.write( "D) -280<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 #515770 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=515770\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Tn = t * r^(n - 1)<BR>\n" );
document.write( "-38 = 3 * r^(7)<BR>\n" );
document.write( "r = -(38/3)^(1/7)\r<BR>\n" );
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document.write( "S=t*(1 - r^n)/(1 - r)<BR>\n" );
document.write( "S=3*(1 - ( -(38/3)^(1/7))^8)/(1 - ( -(38/3)^(1/7)))<BR>\n" );
document.write( "arithmetic<BR>\n" );
document.write( "S = ½(a + an)n<BR>\n" );
document.write( "S = ½(3 + -38)8=-140 <BR>\n" );
document.write( "This is probably meant to be an arithmetic series not GEOMETRIC \r<BR>\n" );
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