document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1103899>Question 1103899</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Taylor deposits $3,500 at the end of year 1, $4,000 at the end of year 3 and $2,500 at the end of year 5. If interest is 6.2% compounded monthly, determine the value at the end of year 7 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #718612 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=718612\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> At the end of year 1, $3500<BR>\n" );
document.write( "at the end of year 3, 3500(1+(.062/12))^24=$3960.79<BR>\n" );
document.write( "puts in 4000 more so $7960.79.<BR>\n" );
document.write( "That is compounded for 24 months to give 7960.79(1+.062/12)^24=$9008.86.<BR>\n" );
document.write( "Puts in 2500 more to have $11508.86<BR>\n" );
document.write( "That is compounded for 24 months to give 11508.86(1+(.062/12)^24=$13024.05<BR>\n" );
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document.write( "rough check--deposited $10000 at 6.2% for 7 years--this is an overestimate: (1+.062/12)^84 is multiplier, which is $15,417. If I use an average of 48 months for the whole amount, it is $12,806.<BR>\n" );
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