document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=567832>Question 567832</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the 15th term in an arithmetic sequence is 43 and the sum of the first 15 terms of the series is 120. Determine the first three terms of the series </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #366729 by </B><A HREF=/tutors/aboutme.mpl?userid=issacodegard><B>issacodegard(60)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=issacodegard><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=issacodegard><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=366729\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Call the terms a_1,a_2,... then S_15=15*(a_1+a_15)/2=120.  So, we solve for a_1,<BR>\n" );
document.write( "a_1=-27.  Then we need to know the common difference, d.  We know that a_n=a_1+(n-1)d.  So, d=(43-(-27))/14=5.  So, a_2=a_1+d=-22, and a_3=a_2+d=-17. </SPAN>\n" );
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