document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171677>Question 1171677</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question: The sum of 40 terms of a certain arithmetic sequence is 430, while the sum of 60 terms is 945. Determine the nth term of the arithmetic sequence.\r<BR>\n" );
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document.write( "What I did so far:  a1 + 39d = 430 ……….Eq(1)<BR>\n" );
document.write( "                                 a1 + 59d = 945 ……… Eq (2)<BR>\n" );
document.write( "             		    Solving for d:   20d = 515  and d = 25.75<BR>\n" );
document.write( "    Therefore the nth term of the arithmetic sequence= a1 +(n  -  1)25.75<BR>\n" );
document.write( "The different answer given in the book is: (n + 1)/2<BR>\n" );
document.write( "What am I missing?<BR>\n" );
document.write( "Please help.<BR>\n" );
document.write( "Thanks<BR>\n" );
document.write( "PLEASE FORGET THIS QUESTION: I USED THE LAST TERM FORMULA AND EQUATED THAT TO SUM!\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #796585 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=796585\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I think you meant to say \"the sum of the first 40 terms is 430\" and \"the sum of the first 60 terms is 945\". Without those \"first\"s in there, the instructions are too vague.\r<BR>\n" );
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document.write( "If that assumption is true, then we use this formula to calculate the sum of the first n terms of an arithmetic sequence<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "where,<BR>\n" );
document.write( "Sn = the sum of the first n terms<BR>\n" );
document.write( "n = number of terms<BR>\n" );
document.write( "a1 = first term<BR>\n" );
document.write( "an = nth term\r<BR>\n" );
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document.write( "If the sum of the first n = 40 terms is 430, then<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "S40 = (40/2)*(a1+a40)<BR>\n" );
document.write( "S40 = 20*(a1+a40)<BR>\n" );
document.write( "430 = 20*(a1+a40)<BR>\n" );
document.write( "430/20 = a1+a40<BR>\n" );
document.write( "21.5 = a1+a40<BR>\n" );
document.write( "a1+a40 = 21.5\r<BR>\n" );
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document.write( "Also, if the sum of the first n = 60 terms is 945, then,<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "S60 = (60/2)*(a1+a60)<BR>\n" );
document.write( "945 = (60/2)*(a1+a60)<BR>\n" );
document.write( "945 = 30*(a1+a60)<BR>\n" );
document.write( "945/30 = a1+a60<BR>\n" );
document.write( "a1+a60 = 31.5\r<BR>\n" );
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document.write( "We have these two equations<BR>\n" );
document.write( "a1+a60 = 31.5<BR>\n" );
document.write( "a1+a40 = 21.5\r<BR>\n" );
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document.write( "Let's subtract the terms straight down<BR>\n" );
document.write( "a1-a1 turns into 0 and goes away<BR>\n" );
document.write( "a60-a40 is unknown<BR>\n" );
document.write( "31.5-21.5 turns into 10\r<BR>\n" );
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document.write( "We can see that <BR>\n" );
document.write( "a60-a40 = 10<BR>\n" );
document.write( "which rearranges to<BR>\n" );
document.write( "a60 = a40+10\r<BR>\n" );
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document.write( "-------------------------------------------\r<BR>\n" );
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document.write( "The n = 40th term is defined as follows<BR>\n" );
document.write( "an = a1 + d(n-1)<BR>\n" );
document.write( "a40 = a1 + d(40-1)<BR>\n" );
document.write( "a40 = a1 + d(39)<BR>\n" );
document.write( "a40 = a1 + 39d\r<BR>\n" );
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document.write( "Similarly, n = 60 leads to<BR>\n" );
document.write( "an = a1 + d(n-1)<BR>\n" );
document.write( "a60 = a1 + d(60-1)<BR>\n" );
document.write( "a60 = a1 + d(59)<BR>\n" );
document.write( "a60 = a1 + 59d\r<BR>\n" );
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document.write( "Now we'll use a60 = a40+10 to get<BR>\n" );
document.write( "a60 = a40+10<BR>\n" );
document.write( "a1 + 59d = a1 + 39d+10<BR>\n" );
document.write( "59d = 39d+10 ............ the a1 terms cancel<BR>\n" );
document.write( "59d-39d = 10<BR>\n" );
document.write( "20d = 10<BR>\n" );
document.write( "d = 10/20<BR>\n" );
document.write( "d = 1/2<BR>\n" );
document.write( "d = 0.5\r<BR>\n" );
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document.write( "-------------------------------------------\r<BR>\n" );
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document.write( "Recall that,<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "We can replace an with a1+d(n-1) and simplify<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "Sn = (n/2)*(a1+a1+d(n-1))<BR>\n" );
document.write( "Sn = (n/2)*(2a1+d(n-1))<BR>\n" );
document.write( "This is an alternative formula to sum the first n terms\r<BR>\n" );
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document.write( "We'll use this alternative formula to help determine a1<BR>\n" );
document.write( "Let's plug in n = 40 and d = 0.5<BR>\n" );
document.write( "Sn = (n/2)*(2a1+d(n-1))<BR>\n" );
document.write( "S40 = (40/2)*(2a1+0.5(40-1))<BR>\n" );
document.write( "430 = 20*(2a1+19.5)<BR>\n" );
document.write( "430/20 = 2a1+19.5<BR>\n" );
document.write( "21.5 = 2a1+19.5<BR>\n" );
document.write( "21.5-19.5 = 2a1<BR>\n" );
document.write( "2 = 2a1<BR>\n" );
document.write( "2a1 = 2<BR>\n" );
document.write( "a1 = 2/2<BR>\n" );
document.write( "a1 = 1\r<BR>\n" );
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document.write( "Or we could use n = 60 with d = 0.5<BR>\n" );
document.write( "Sn = (n/2)*(2a1+d(n-1))<BR>\n" );
document.write( "S60 = (60/2)*(2a1+0.5(60-1))<BR>\n" );
document.write( "945 = 30*(2a1+29.5)<BR>\n" );
document.write( "945/30 = 2a1+29.5<BR>\n" );
document.write( "31.5 = 2a1+29.5<BR>\n" );
document.write( "31.5-29.5 = 2a1<BR>\n" );
document.write( "2 = 2a1<BR>\n" );
document.write( "2a1 = 2<BR>\n" );
document.write( "a1 = 2/2<BR>\n" );
document.write( "a1 = 1<BR>\n" );
document.write( "We end up with the same starting term.\r<BR>\n" );
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document.write( "The last thing to do is to plug a1 = 1 and d = 0.5 into the nth term formula below<BR>\n" );
document.write( "an = a1 + d(n-1)<BR>\n" );
document.write( "an = 1 + 0.5(n-1)<BR>\n" );
document.write( "an = 1 + 0.5n-0.5<BR>\n" );
document.write( "an = 0.5n + 1-0.5<BR>\n" );
document.write( "an = 0.5n + 0.5<BR>\n" );
document.write( "an = 0.5(n + 1)<BR>\n" );
document.write( "an = (1/2)(n + 1)<BR>\n" );
document.write( "an = (n + 1)/2\r<BR>\n" );
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document.write( "-------------------------------------------\r<BR>\n" );
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document.write( "As a way to check, plug in n = 40<BR>\n" );
document.write( "an = (n + 1)/2<BR>\n" );
document.write( "a40 = (40 + 1)/2<BR>\n" );
document.write( "a40 = 20.5<BR>\n" );
document.write( "Then note how<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "S40 = (40/2)*(1+20.5)<BR>\n" );
document.write( "S40 = 430<BR>\n" );
document.write( "which matches with the instructions\r<BR>\n" );
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document.write( "Also, plug in n = 60 <BR>\n" );
document.write( "an = (n + 1)/2<BR>\n" );
document.write( "a60 = (60 + 1)/2<BR>\n" );
document.write( "a60 = 30.5<BR>\n" );
document.write( "which means,<BR>\n" );
document.write( "Sn = (n/2)*(a1+an)<BR>\n" );
document.write( "S60 = (60/2)*(1+30.5)<BR>\n" );
document.write( "S60 = 945<BR>\n" );
document.write( "and that matches as well. The answer has been confirmed.\r<BR>\n" );
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document.write( "A longer winded way of checking the answer is to list the first 40 terms of the sequence (which is now possible since we know a1 = 1 and d = 0.5), then add up said terms. Do the same for the first 60 terms as well. This is not recommended unless you are using computer software to make the process relatively painless.\r<BR>\n" );
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document.write( "-------------------------------------------\r<BR>\n" );
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document.write( "Answer: (n + 1)/2\r<BR>\n" );
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