document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1027240>Question 1027240</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the first three terms of a geometric progression are 100,90 and 81.\r<BR>\n" );
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document.write( "Find out the common ratio?<BR>\n" );
document.write( "find the sum to infinity of its terms?\r<BR>\n" );
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document.write( "can I be explained step by step how to figure this out  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #642501 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=642501\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a geometric progression has the form An = A1 * r^(n-1)\r<BR>\n" );
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document.write( "An is the nth term.<BR>\n" );
document.write( "A1 is the first term.<BR>\n" );
document.write( "n is the number of terms.\r<BR>\n" );
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document.write( "your problem has the sequence 100, 90, 81, .....\r<BR>\n" );
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document.write( "the common ratio is .9 because 100 * .9 = 90 and 90 * .9 = 81.\r<BR>\n" );
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document.write( "the formula of An = A1 * r^(n-1) therefore becomes An = 100 * .9^(n-1)\r<BR>\n" );
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document.write( "for example, the third term, which you alreay know is .81, would be calculated using this forula as A3 = 100 * .9^(2.\r<BR>\n" );
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document.write( "you would then get A3 = 81.\r<BR>\n" );
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document.write( "this agrees with what you already know so the formula is confirmed as good.\r<BR>\n" );
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document.write( "you want to know the sum of this progression as n approaches infinity.\r<BR>\n" );
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document.write( "the formula for the sum of the geometric progression as n approaches infinity is:\r<BR>\n" );
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document.write( "Sum of geometric progression as n approaches infinity = A1 * 1 / (1-r)\r<BR>\n" );
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document.write( "for your progression, that would be 100 * 1/(1-.9) = 100 / .1 = 1000\r<BR>\n" );
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document.write( "the following reference discusses geometric progressions.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.purplemath.com/modules/series5.htm\" target = \"_blank\">http://www.purplemath.com/modules/series5.htm</a>\r<BR>\n" );
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document.write( "i used excel to analyze whether this was true or not.<BR>\n" );
document.write( "it turns out to be true.<BR>\n" );
document.write( "after about the 160th term, the sum reached 1000 and didn't go over.\r<BR>\n" );
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document.write( "the formula for the sum of the geometric progression to the nth term is:\r<BR>\n" );
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document.write( "sum to the nth term = A1 * (1 - r^n) / (1-r)\r<BR>\n" );
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document.write( "for the 160th term, that formula would become sum = 100 * (1 - .9^160) / (1-.9) which becomes sum = 100 * .9999999523/.10 which becomes equal to 999.9999523.\r<BR>\n" );
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document.write( "that's very close to 1000.\r<BR>\n" );
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document.write( "it gets closer as n gets higher, but never goes over 1000.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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