document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1133111>Question 1133111</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the numbers 4,7,10,13,16 where each number is three greater than the number preceding it, are written in order in a book, one hundred to a page. The first group of one hundred numbers begins on page 526. on which page number 2005 be located?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #750294 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=750294\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the formula for the nth term in an arithmetic progression is<BR>\n" );
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document.write( "a(n) = a(0) +d(n-1), where a(0) is the first term, d is the common difference<BR>\n" );
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document.write( "for this problem, a(n) is 2005, a(0) is 4 and d is 3<BR>\n" );
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document.write( "we want to find n for the term 2005<BR>\n" );
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document.write( "2005 = 4 +3(n-1)<BR>\n" );
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document.write( "3(n-1) = 2001<BR>\n" );
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document.write( "3n -3 = 2001<BR>\n" );
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document.write( "3n = 2004<BR>\n" );
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document.write( "n = 668<BR>\n" );
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document.write( "there are 100 numbers per page, then<BR>\n" );
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document.write( "668/100 = 6 with a remainder of 68<BR>\n" );
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document.write( "2005 is found on page 526 + 7 = 533<BR>\n" );
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