document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=831211>Question 831211</A>: <I><SPAN STYLE=\"word-break: break-all;\"> given the arithmetic sequence 58, 73, 88 determine:<BR>\n" );
document.write( "-the general term, tn<BR>\n" );
document.write( "-the value of the tenth term, t10<BR>\n" );
document.write( "-the sum of the first ten terms, S10 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #501213 by </B><A HREF=/tutors/aboutme.mpl?userid=Elomeht><B>Elomeht(22)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Elomeht><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=501213\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We are dealing here with an arithmetic progression where the the first term, a, is 58 and the common difference, d, is 15.\r<BR>\n" );
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document.write( "1.  The general formula for the nth term of an arithmetic progression is:\r<BR>\n" );
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document.write( "    a + (n - 1)d\r<BR>\n" );
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document.write( "   Substituting for a and d, we get 58 + 15(n - 1)\r<BR>\n" );
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document.write( "2.  When n = 10, we have 58 + 15 times 9 = 58 + 135 = 193\r<BR>\n" );
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document.write( "3.  The sum of the first n terms is given by the formula:<BR>\n" );
document.write( "    (n/2)[2a + (n - 1)d]<BR>\n" );
document.write( "    Substituting for a, n and d, we get:\r<BR>\n" );
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document.write( "    (10/2)[2 times 58 + 15 times 9] = 5[116 + 135] = 5 times 251 = 1255 </SPAN>\n" );
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