document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1023887>Question 1023887</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Sam bought a home with a 30 year mortgage of $200,000 at an annual interest rate of 4.25%.  Assuming Sam makes his monthly payment over the term of the loan, what will be the total amount of his payments?  Also what will be his payment per month if he pays the same amount monthly?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #692433 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=692433\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To amortize a loan, the formula is P=L[c(1+c)^n]/[(1+c)^n-1] where P is the payment,L is the original loan, c is the interest rate per period, and n is the total number of payments. So:<BR>\n" );
document.write( "P=200000[(.0425/12)(1+(.0425/12))^360))/[(1+(.0425/12))^360 -1]<BR>\n" );
document.write( "P=2529.2100430477458909800905736667/2.5706494725379941990307161040001<BR>\n" );
document.write( "P=$983.88/month<BR>\n" );
document.write( "360 x 983.88=$354196.80 total payments over 30 years. &#9786;&#9786;&#9786;&#9786; </SPAN>\n" );
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