document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1085372>Question 1085372</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the first numbers p-1,2p-2 and 3p-1 are the first three terms of a Gp where p>0 find <BR>\n" );
document.write( "(i) the sum to infinity </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #699425 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=699425\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=\"black\" face=\"times\" size=\"3\">GP means \"geometric progression\" which is another way to say \"geometric sequence\".\r<BR>\n" );
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document.write( "Term 1 = p-1<BR>\n" );
document.write( "Term 2 = 2p-2<BR>\n" );
document.write( "Term 3 = 3p-1\r<BR>\n" );
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document.write( "The common ratio r is the ratio of adajcent terms.<BR>\n" );
document.write( "Specifically it's the result of dividing any term (but the first one) over its previous term.<BR>\n" );
document.write( "The basic template is\r<BR>\n" );
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document.write( "common ratio = (any term but the first term)/(previous term)\r<BR>\n" );
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document.write( "Using that template, we can divide term 2 over term 1 to get\r<BR>\n" );
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document.write( "r = (term 2)/(term 1)<BR>\n" );
document.write( "r = (2p-2)/(p-1)<BR>\n" );
document.write( "r = (2(p-1))/(p-1)<BR>\n" );
document.write( "r = 2\r<BR>\n" );
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document.write( "The (p-1) terms cancel out leaving r = 2, which is the common ratio. \r<BR>\n" );
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document.write( "Using the same template, we can divide term 3 over term 2 to get\r<BR>\n" );
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document.write( "r = (term 3)/(term 2)<BR>\n" );
document.write( "r = (3p-1)/(2p-2)\r<BR>\n" );
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document.write( "Plug in r = 2 and solve for p.\r<BR>\n" );
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document.write( "r = (3p-1)/(2p-2)<BR>\n" );
document.write( "2 = (3p-1)/(2p-2)<BR>\n" );
document.write( "2(2p-2) = [(3p-1)/(2p-2)]*(2p-2)<BR>\n" );
document.write( "2(2p-2) = 3p-1<BR>\n" );
document.write( "4p-4 = 3p-1<BR>\n" );
document.write( "4p-3p = -1+4<BR>\n" );
document.write( "p = -1+4<BR>\n" );
document.write( "p = 3\r<BR>\n" );
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document.write( "We know the value of p is p = 3, so make substitutions to get\r<BR>\n" );
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document.write( "Term 1 = p-1 = 3-1 = 2<BR>\n" );
document.write( "Term 2 = 2p-2 = 2*3-2 = 4<BR>\n" );
document.write( "Term 3 = 3p-1 = 3*3-1 = 8\r<BR>\n" );
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document.write( "The first three terms are\r<BR>\n" );
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document.write( "Term 1 = 2<BR>\n" );
document.write( "Term 2 = 4<BR>\n" );
document.write( "Term 3 = 8\r<BR>\n" );
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document.write( "So the GP is 2,4,8,...\r<BR>\n" );
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document.write( "This GP goes on forever and there is no limit to how big the terms grow. <BR>\n" );
document.write( "This means that the sum to infinity is not a fixed number. <BR>\n" );
document.write( "The sum will be infinity implying that it keeps growing forever (much like the terms do).<BR>\n" );
document.write( "For this current problem, we cannot find the sum to infinity. \r<BR>\n" );
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document.write( "The only way to have the sum be a finite number, we need to have the common ratio r be between 0 and 1. <BR>\n" );
document.write( "The value of r can't equal 0. The value of r can't equal 1.<BR>\n" );
document.write( "Put another way we need <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+%3C+r+%3C+1 ALIGN=MIDDLE  ALT=\"0+%3C+r+%3C+1\"   DISABLED_style= \"max-width:90%; height:auto;\" > for the series to converge. <BR>\n" );
document.write( "If <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+%3E=+1 ALIGN=MIDDLE  ALT=\"r+%3E=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >, then the series diverges.</font> </SPAN>\n" );
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