document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1022028>Question 1022028</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the sum of the first 10 terms of geometric series starting at 10 and decreasing by a factor of 2 each time?<BR>\n" );
document.write( "A 5 - 5/2^8<BR>\n" );
document.write( "B 10 - 5/2^8<BR>\n" );
document.write( "C 15 - 5/2^8<BR>\n" );
document.write( "D 20 - 5/2^8 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #637709 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=637709\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the sum of a geometric series is Sn = a(1 - r^n) / (1 - r) where a is the first term, r is the common ratio, and n is n consecutive terms<BR>\n" );
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document.write( "S10 = 10(1 - (1/2)^10) / ( 1 - (1/2))<BR>\n" );
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document.write( "S10 = 10(1 - (1/(2^10))) / (1/2)<BR>\n" );
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document.write( "divide by 1/2 means multiply by 2<BR>\n" );
document.write( "S10 = 20(1 - (1 /2^10)<BR>\n" );
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document.write( "20 = 2^2 * 5<BR>\n" );
document.write( "S10 = 20 - (((2^2)*5)/ 2^10)<BR>\n" );
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document.write( "S10 = 20 - (5 / 2^8)<BR>\n" );
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document.write( "Answer is D<BR>\n" );
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