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<div id="main-column" class="span9"> <article id="1b-arithmetic-sequences-series-sn" style="margin-top: 16px;">
<h1 class="section-title">1B. Arithmetic sequences & series (SN)</h1>
<ul class="breadcrumb"><li><a title="Home" href="../../../index.html"><i class="fa fa-home"></i></a><span class="divider">/</span></li><li><span class="gray">1. Number & Algebra</span><span class="divider">/</span></li><li><span class="active">1B. Arithmetic sequences & series (SN)</span></li></ul>
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<h3>∼ Student Notes ∼</h3><h3><img alt="" height="152" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/arith-img01.jpg" style="float: right;" width="272">Section 1B. Arithmetic sequences & series</h3><p style="margin-left: 40px;"><strong>Sub-sections:</strong><br><a class="scroll-to" data-target="general"><u>1B.1</u> General term of an arithmetic sequence</a><br><a class="scroll-to" data-target="graph"><u>1B.2</u> Graph of an arithmetic sequence</a><br><a class="scroll-to" data-target="series"><u>1B.3</u> Arithmetic series</a><br><a class="scroll-to" data-target="sigma"><u>1B.4</u> Sigma notation</a><br><a class="scroll-to" data-target="apps"><u>1B.5</u> Applications of arithmetic sequences & series</a><br><a class="scroll-to" data-target="summary"><u>1B.6</u> Summary</a></p><div class="blueBg"><p><span style="font-size:12.0pt;line-height:115%;
font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;mso-ansi-language:
EN-GB;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">■</span> syllabus content covered in this section <span style="font-size:12.0pt;line-height:115%;
font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;mso-ansi-language:
EN-GB;mso-fareast-language:EN-US;mso-bidi-language:AR-SA">■</span></p><section class="tib-hiddenbox"><p><strong>SL 1.2* </strong>Arithmetic sequences and series; use of the formulae for the <span class="math-tex">\(n\)</span><sup>th</sup> term and the sum of the first<em> </em><span class="math-tex">\(n\)</span> terms of the sequence; use of sigma notation for sums of arithmetic sequences; applications (e.g. simple interest); interpretation and prediction where a model is not perfectly arithmetic in real life.</p></section></div><hr><h4><strong><a class="anchor" id="general" name="general"> </a>1B.1 General term of an arithmetic sequence</strong></h4><p>Imagine that someone is willing to pay you for a job that lasts for exactly 30 days - and that they ask how you would like to be paid for the job. Let's consider the following two payment plans that you could propose to your prospective employer.</p><p><u><strong>Plan A</strong></u>: You get paid $1 for the first day and for each day after that you get paid $3 more than the previous day. So, the payments for the first week is the <strong>sequence </strong><span class="math-tex">\(1,\;4,\;7,\;10,\;13,\;16,\;19\)</span> . Your total (cumulative) pay for the first week is given by the sum of the <strong>series</strong> <span class="math-tex">\(1 + 4 + 7 + 10 + 13 + 16 + 19\)</span> which is <span class="math-tex">\($70\)</span>.</p><p><u><strong>Plan B</strong></u>: You get paid just $0.01 (one cent) for the first day and for each day after that you get paid two times what you were paid the previous day. So, the payments for the first week is the <strong>sequence </strong><span class="math-tex">\(0.01,\;0.02,\;0.04,\;0.08,\;0.16,\;0.32,\;0.64\)</span> . Your cumulative pay for the first week is the sum of the <strong>series </strong><span class="math-tex">\(0.01 + 0.02 + 0.04 + 0.08 + 0.16 + 0.32 + 0.64\)</span> which is <span class="math-tex">\($1.27\)</span>.</p><p>Which payment plan do you think is best for you? Which payment plan do you think your employer would choose? After one week, Plan A will earn you about 55 times what you would earn with Plan B (<span class="math-tex">\(\frac{{70}}{{1.27}} \approx 55.12\)</span>); and after two weeks, you will have earned <span class="math-tex">\($287\)</span> with Plan A, but only <span class="math-tex">\($163.83\)</span> with Plan B. Answering these questions and a closer consideration of Plan B must wait until the next section (1C). For now, let's take a closer look at Plan A.</p><div class="row-fluid"><div class="span6 col-left"><div><p>Recall from the previous section (<strong>1A</strong>), that a sequence where each term (after the first term) is determined by adding a constant to the previous term is called an <strong>arithmetic sequence</strong>. Clearly, the list of daily payments for Plan A is an arithemtic sequence. We can make use of the <strong>recursive </strong>rule for Plan A, "start with <span class="math-tex">\($1\)</span> for the 1<sup>st</sup> day; then each subsequent day's pay is <span class="math-tex">\($3\)</span> added to the previous day's pay", to quickly construct the sequence of 30 payments using Excel (see video at right; no sound). We can see from the spreadsheet results that with Plan A you would make <span class="math-tex">\($58\)</span> on the 20th day. How can we calculate this without having to find what you would make on each of the days before the 20<sup>th</sup> day? In other words, we'd like to have an <strong>explicit </strong>rule (formula) for finding the <span class="math-tex">\(n\)</span><sup>th</sup> term of the sequence. As Figure 1 shows, your payment of <span class="math-tex">\($13\)</span> on the 5<sup>th</sup> day can be calculated (recursively) by adding <span class="math-tex">\($3\)</span> to the 1<sup>st</sup> day's payment four times. To get from the 1<sup>st</sup> term (<span class="math-tex">\(n = 1\)</span>) in a sequence to the 5<sup>th</sup> term (<span class="math-tex">\(n = 5\)</span>) in a sequence, the common difference <span class="math-tex">\(d\)</span> is added 4 times (Figure 2). To quickly calculate your pay on the 20<sup>th</sup> day, add 19 payments of <span class="math-tex">\($3\)</span> to <span class="math-tex">\($1\)</span> (1<sup>st</sup> day's pay): <span class="math-tex">\(\$ 1 + \left( {20 - 1} \right) \cdot \$ 3 = 19 \cdot \$ 3 = \$ 58\)</span>.</p></div></div><div class="span6"><div><p style="text-align: center;"><iframe frameborder="0" height="391" scrolling="no" src="https://player.vimeo.com/video/437431996" width="160"></iframe></p></div></div></div><div class="polaroid-left"><img src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/arith-img1a.jpg" style="margin:8px 0"><div class="caption">Figure 1</div></div><div class="polaroid-left"><img src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/arith-img2a.jpg" style="margin:8px 0"><div class="caption">Figure 2</div></div><hr class="hidden"><p>In general, to find the <span class="math-tex">\(n\)</span><sup>th</sup> term of an arithmetic sequence add <span class="math-tex">\(\left( {n - 1} \right)\)</span> times the common difference <span class="math-tex">\(d\)</span> to the first term.</p><div class="pinkBg"><h4><span class="math-tex">\(n\)</span><sup>th</sup> term (general term) of an arithmetic sequence</h4><p>The <span class="math-tex">\(n\)</span><sup>th</sup> term, <span class="math-tex">\({u_n}\)</span>, of an <strong>arithmetic sequence</strong> with first term <span class="math-tex">\({u_1}\)</span> (initial term) and common difference <span class="math-tex">\(d\)</span> is given by the following <strong>explicit formula</strong> (in the <a href="/media/ib/mathanalysis/analysis/basics/aa-formula-bklet-v1.2.pdf" target="_blank">Analysis & Approaches formula booklet</a>):</p><p><span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\)</span></p><p><u>note</u>: Although <span class="math-tex">\(n\)</span> often represents the position (1<sup>st</sup>, 4<sup>th</sup>, etc) of the term in a sequence, it is also common to use other letters such as <span class="math-tex">\(r\)</span> and <span class="math-tex">\(i\)</span> instead of <span class="math-tex">\(n\)</span>. Also, it is common to use letters other than <span class="math-tex">\(u\)</span> such as <span class="math-tex">\(a\)</span> or <span class="math-tex">\(t\)</span> to represent the terms of a sequence.</p></div><h5>Example 1</h5><p><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/leon2_img.jpg" style="float: left; width: 115px; height: 126px;">Leon feels that he is spending too much time playing video games. He estimates that he spends 300 minutes each day playing video games. Leon decides to commit to a 'cutback program' where he will reduce the time spent on video games by 7 minutes each day (his favourite number is 7 because he was born on 7-7-2007).<br>(a) If Leon plays video games for 300 minutes on the first day of his program, then how many minutes will Leon play video games on the 30<sup>th</sup> day of his program?<br>(b) Leon decides he will stop his program on the first day that he plays video game for less than one hour. On what day does he stop his 'cutback program'?</p><div class="box"><p><strong>Example 1 solution</strong>:</p><section class="tib-hiddenbox"><p>(a) The amount of time that Leon plays video games on each day of his program is an arithmetic sequence where <span class="math-tex">\({u_1} = 300\)</span> and <span class="math-tex">\(d = - \,7\)</span>. Using the formula <span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\)</span>, his time spent playing video games on the 30<sup>th</sup> day is given by: <span class="math-tex">\({u_{30}} = 300 + \left( {30 - 1} \right)\left( { - 7} \right) = 300 - 203 = 97\)</span> . If Leon sticks precisely to his program, he will play video games for 97 minutes on the 30<sup>th</sup> day.</p><p>(b) To find the first day that Leon plays video games for less than one hour, solve the equation <span class="math-tex">\({u_n} = 300 + \left( {n - 1} \right)\left( { - 7} \right) = 60\)</span>. Solving for <span class="math-tex">\(n\)</span> gives <span class="math-tex">\(300 - 7n + 7 = 60\;\; \Rightarrow \;\;7n = 247\;\; \Rightarrow \;\;n = \frac{{247}}{7} \approx 35.29\)</span><br>On the 35<sup>th</sup> day, Leon will not quite be below 60 minutes: <span class="math-tex">\(300 + \left( {35 - 1} \right)\left( { - 7} \right) = 62\)</span>. Hence, <span class="math-tex">\(n=36\)</span>; and Leon will stop his program on the 36<sup>th</sup> day when he will play video games for 55 minutes (<span class="math-tex">\(62-7\)</span>).</p></section></div><p><u>Note</u>: There is a very crude but simple way to use a GDC to generate the sequence to answer the two questions in Example 1. Watch the video below (with audio description).</p><p style="text-align: center;"><iframe frameborder="0" height="225" scrolling="no" src="https://player.vimeo.com/video/437913075" width="300"></iframe></p><p>Generally speaking, it is best to apply the explicit formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term for answering a question like Example 1 rather than repeatedly applying the recursive rule on a GDC. Although it's nice to see the terms listed one after another, it is easy to make an error when trying to keep track of what number term (position in the sequence) was just calculated. Also, writing out a long list of terms as your work for answering an exam question would be frowned upon. And, of course, this would not be possible on a Paper 2 exam question because you would not have a GDC with you.</p><hr><h4><a class="anchor" id="graph" name="graph"> </a>1B.2 Graph of an arithmetic sequence</h4><p>A better use of a GDC is to make a graph to illustrate the behaviour of a sequence. The formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term in Example 1, <span class="math-tex">\({u_n} = 300 + \left( {n - 1} \right)\left( { - 7} \right)\)</span>, can be simplified to <span class="math-tex">\({u_n} = - 7n + 307\)</span>. Thus, the formula for <span class="math-tex">\({u_n}\)</span> is a <strong>linear function</strong> where <span class="math-tex">\(n\)</span>, the <em>input</em>, is a positive integer, and <span class="math-tex">\({u_n}\)</span> is the <em>output</em>. The graph of <span class="math-tex">\({u_n} = - 7n + 307\)</span> will <u>not</u> be a smooth continuous curve but a discrete set of points: <br><span class="math-tex">\(\left( {1,300} \right),\left( {2,293} \right),\left( {3,286} \right), \ldots ,\left( {35,62} \right),\left( {36,55} \right)\)</span></p><div class="row-fluid"><div class="span6 col-left"><div><p>The video at right (no sound) shows a way to graph the arithmetic sequence of 36 terms from Example 1 on a <strong>TI-84</strong> calculator. This is done by creating two sequences: first a list (stored in L1) of the positive integers from 1 to 36 for the set of values for the input <span class="math-tex">\(n\)</span> (number/position of the term), and a second sequence for a list (stored in L2) of the values for the ouput <span class="math-tex">\({u_n}\)</span>. Then a Stat Plot is set up that graphs L1 vs L2 (both lists) on the <em>x</em>-<em>y</em> coordinate plane where the values for <span class="math-tex">\(n\)</span> in list L1 are the <em>x</em>-coordinates and the values for <span class="math-tex">\({u_n}\)</span> in list L2 are the <em>y</em>-coordinates. The discrete plot (graph) clearly illustrates that the points <span class="math-tex">\(\left( {n,{u_n}} \right)\)</span> generated by the formula <span class="math-tex">\({u_n} = - 7n + 307\)</span> are all on a line; i.e. a linear function. The gradient of the line is <span class="math-tex">\( - 7\)</span> which corresponds to the common difference <span class="math-tex">\(d\)</span> for the sequence. The 36 terms in the sequence for Example 1 show a constant decrease (<span class="math-tex">\(d = - 7\)</span>), or <em>linear decay</em>.</p></div></div><div class="span6"><div><p><iframe frameborder="0" height="256" scrolling="no" src="https://player.vimeo.com/video/438006304" width="350"></iframe></p></div></div></div><hr class="hidden"><p><iframe align="left" frameborder="0" height="264" scrolling="no" src="https://player.vimeo.com/video/438036518" width="352"></iframe></p><div class="box"><p>← <u>video</u>: plotting sequence from Example 1 on the <strong>TI-Nspire</strong> where points are <span class="math-tex">\(\left( {n,{\textrm{time}}} \right)\)</span></p></div><div class="greenBg"><h4>Graph (plot) of an arithmetic sequence</h4><p>The explicit <strong>formula </strong>for the <span class="math-tex">\(n\)</span><sup>th</sup> term of an <strong>arithmetic sequence</strong> is a <strong>linear function</strong> whose plot is a <strong>set of discrete points</strong>, <span class="math-tex">\(\left( {n,{u_n}} \right)\)</span>, that all lie on a line with a gradient of <span class="math-tex">\(d\)</span>, the common difference of the sequence. The terms in an arithmetic sequence display <strong>linear growth</strong> (positive <span class="math-tex">\(d\)</span>) or <strong>linear decay</strong> (negative <span class="math-tex">\(d\)</span>).</p></div><hr class="hidden"><p><strong>Modelling with an arithmetic sequence</strong></p><p><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/lilly_img.jpg" style="float: left; width: 97px; height: 131px;">Recall Leon's 'cutback program' in Example 1. Perhaps Leon did reduce his time playing video games by <em>exactly </em>7 minutes each day, but it would be a difficult to be that precise. Lilly, a friend of Leon's, has decided to <em>increase </em>the time she plays video games each day because she is preparing for an e-sports competition. The following list is her time (in minutes) playing video games each day for the first 10 days (in order) of her 'buildup program': <span class="math-tex">\(100,112,121,132,142,154,165,175,188,198\)</span>. It's clear that Lilly did not increase her time by exactly the same amount each day. For the ten terms in this sequence, the nine <em>differences</em> from one term to the next are: <span class="math-tex">\(12,9,11,10,12,11,10,13,10\)</span>. The average difference is <span class="math-tex">\(10.\bar 8\)</span> minutes. The sequence of Lilly's times for the first 10 days of her program can be modelled (approximated) by an arithmetic sequence with a common difference of 11. We can use the linear function from the formula for the general term of the 'model' sequence to approximate Lilly's video game time on future days in her program (assuming her program continues similarly to the first 10 days).</p><p><strong>Example 2</strong>: Lilly decides to stop her 'buildup program' after 3 weeks. For how many minutes will Lilly play video games on the last day (21<sup>st</sup>) of her program?</p><div class="box"><p><strong>Example 2 solution:</strong></p><section class="tib-hiddenbox"><p>The arithmetic sequence that is a reasonable model for the sequence of daily times for Lilly has <span class="math-tex">\({u_1} = 100\)</span> and <span class="math-tex">\(d = 11\)</span>. Hence, a reasonable estimate for the amount of time that Lilly will play video games on the 21<sup>st</sup> day can be calcuated using the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of an arithmetic sequence which will be a linear function:<br><span class="math-tex">\({u_n} = 100 + \left( {n - 1} \right) \cdot 11\;\;\; \Rightarrow \;\;\;{u_n} = 11n + 89\)</span> Evaluate the function for <span class="math-tex">\(n = 21\)</span>: <span class="math-tex">\({u_n} = 11\left( {21} \right) + 89 = 320\)</span><br>Hence, the prediction is that Lilly will play video games for approximately 320 minutes on the last day of her 3-week program.</p></section></div><p>The images below (from a TI-Nspire) illustrate a graphical approach to modelling Lilly's daily times. The middle image shows a scatter plot of Lilly's times for the first 10 days of her program; and the image on the right includes a graph of the linear (model) function <span class="math-tex">\(f\left( x \right) = 11x + 89\)</span> that estimates Lilly's time on the <span class="math-tex">\(x\)</span><sup>th</sup> day of her program. A trace on the graph shows that <span class="math-tex">\(f\left( {21} \right) = 320\)</span>.</p><p><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/nspire-lilly-data-img.jpg" style="width: 110px; height: 235px; float: left; margin: 0px 10px;"><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/nspire-lilly-graph1-img.jpg" style="margin: 0px 10px; float: left; width: 250px; height: 186px;"><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/nspire-lilly-graph2-img.jpg" style="margin: 0px 10px; float: left; width: 250px; height: 185px;"></p><hr class="hidden"><hr><h4><a class="anchor" id="series" name="series"> </a>1B.3 Arithmetic series (finding the sum of an arithmetic sequence)</h4><p><strong><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/scooter.jpg" style="margin: 0px 5px; float: right; width: 112px; height: 122px;">Example 3</strong></p><p>From a stationary position, an electric scooter is accelerating in a straight line at a constant rate such that in the 1<sup>st</sup> second it moves 0.5 metres, in the 2<sup>nd</sup> second it moves 2 metres, in the 3<sup>rd</sup> second it moves 3.5 metres, and in the 4<sup>th</sup> second it moves 5 metres. If this pattern continues, then what is the total distance travelled by the scooter in the first 10 seonds?</p><div class="box"><p><strong>Example 3 solution:</strong></p><section class="tib-hiddenbox"><p>The pattern of the scooter's acceleration is that for the 2<sup>nd</sup> second onwards it travels 1.5 metres further than the previous second. Hence, the list of distances travelled for each second is an arithmetic sequence with <span class="math-tex">\({u_1} = 0.5\)</span> and <span class="math-tex">\(d = 1.5\)</span> . The sum of the first 10 terms of this sequence (which we can label as <span class="math-tex">\({S_{10}}\)</span>) is the <strong>finite arithmetic series</strong> <span class="math-tex">\({S_{10}} = 0.5 + 2 + 3.5 + 5 + 6.5 + 8 + 9.5 + 11 + 12.5 + 14\)</span>. Calculating <span class="math-tex">\({S_{10}}\)</span> is straightforward, if a bit tedious, but if the sum had been for the first 25 seconds, an efficient algebraic technique would be very useful.</p></section></div><p>There is an efficient way to find the sum of an arithemtic series that is based on a clever approach used by the famous mathematician <strong>Carl Friedrich Gauss</strong> (1777-1855) when he was a very young student (see <a href="http://www.youtube.com/watch?v=j-EB1O-vRS4" target="_blank">video</a> from the <a href="https://artofproblemsolving.com/" target="_blank">Art of Problem Solving</a> site).</p><p>For the 10 terms in the series <span class="math-tex">\({S_{10}} = 0.5 + 2 + 3.5 + 5 + 6.5 + 8 + 9.5 + 11 + 12.5 + 14\)</span>, add the 1<sup>st</sup> and 10<sup>th</sup> terms (<span class="math-tex">\(0.5 + 14 = 14.5\)</span>), and add the 2<sup>nd</sup> and 9<sup>th</sup> terms (<span class="math-tex">\(2 + 12.5 = 14.5\)</span>), and add the 3<sup>rd</sup> and 8<sup>th</sup> terms (<span class="math-tex">\(3.5 + 11 = 14.5\)</span>), the 4<sup>th</sup> and 7<sup>th</sup> terms (<span class="math-tex">\(5 + 9.5 = 14.5\)</span>), and, finally, the 5<sup>th</sup> and 6<sup>th</sup> terms (<span class="math-tex">\(6.5 + 8 = 14.5\)</span>). Each of these pairs of terms adds up to <span class="math-tex">\(14.5\)</span>, and the number of pairs is 5 (half of the total # of terms). Hence, the sum of this series is <span class="math-tex">\({S_{10}} = \frac{{10}}{2}\left( {14.5} \right) = 72.5\)</span>. The total distance travelled by the scooter in the first 10 seconds is <span class="math-tex">\(72.5\)</span> metres.</p><div class="yellowBg"><p><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/seq-series/gauss-img2.jpg" style="float: right; width: 90px; height: 110px; margin: 0px;">An excerpt from Gauss' <a href="https://mathshistory.st-andrews.ac.uk/Biographies/Gauss/" target="_blank">biography</a> on the <a href="https://mathshistory.st-andrews.ac.uk/" target="_blank">MacTutor History of Mathematics Archive</a> (University of St Andrews):</p><p><em>At the age of seven, Carl Friedrich Gauss started elementary school, and his potential was noticed almost immediately. His teacher … and his assistant … were amazed when Gauss summed the integers from 1 to 100 instantly by spotting that the sum was 50 pairs of numbers each pair summing to 101.</em></p></div><p>The technique used by Gauss, and in the solution for Example 3, can be applied to develop a <strong>general formula for the sum of an arithmetic series</strong>. The key aspect of the technique is adding pairs of terms: first term plus last term, and second term plus next-to-last term (<span class="math-tex">\({u_2} + {u_{n - 1}}\)</span>), and <span class="math-tex">\({u_3} + {u_{n - 2}}\)</span>, and so on.</p><p>Let <span class="math-tex">\({S_n}\)</span> be the sum of an arithmetic series with <span class="math-tex">\(n\)</span> terms: <span class="math-tex">\({S_n} = {u_1} + {u_2} + {u_3} + \cdots + {u_{n - 2}} + {u_{n - 1}} + {u_n}\)</span></p><p>... write the series with the order of the terms reversed: <span class="math-tex">\({S_n} = {u_n} + {u_{n - 1}} + {u_{n - 2}} + \cdots + {u_3} + {u_2} + {u_1}\)</span></p><p>... add these two expressions: <span class="math-tex">\(2{S_n} = \left( {{u_1} + {u_n}} \right) + \left( {{u_2} + {u_{n - 1}}} \right) + \left( {{u_3} + {u_{n - 2}}} \right) + \cdots + \left( {{u_1} + {u_n}} \right) + \left( {{u_2} + {u_{n - 1}}} \right) + \left( {{u_3} + {u_{n - 2}}} \right)\)</span></p><p>... it follows that: <span class="math-tex">\(2{S_n} = n\left( {{u_1} + {u_n}} \right)\)</span>; which leads to the general formula <span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\)</span></p><p>This mirrors the working shown in the solution of Example 3 where the sum of the series <span class="math-tex">\(0.5 + 2 + 3.5 + 5 + 6.5 + 8 + 9.5 + 11 + 12.5 + 14\)</span> is found with the calculation <span class="math-tex">\({S_{10}} = \frac{{10}}{2}\left( {14.5} \right)\)</span> which equals <span class="math-tex">\(72.5\)</span></p><div class="pinkBg"><h4>Sum of an arithmetic series with <em>n</em> terms (version 1)</h4><p>The sum of <span class="math-tex">\(n\)</span> terms, <span class="math-tex">\({S_n}\)</span>, of an <strong>arithmetic series</strong> with first term <span class="math-tex">\({u_1}\)</span> and last term <span class="math-tex">\({u_n}\)</span> is given by the following<strong> </strong>formula (in the <a href="/media/ib/mathanalysis/analysis/basics/aa-formula-bklet-v1.2.pdf" target="_blank">Analysis & Approaches formula booklet</a>):</p><p><span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\)</span></p></div><p>What if we want to find the sum of a series but we do not know the last term?</p><p><strong>Example 4</strong></p><p>Find the sum of the first 25 terms of the following arithmetic series: <span class="math-tex">\(13 + 9 + 5 + 1 - 3 - 7 - \; \cdots \)</span></p><div class="box"><p><strong>Example 4 solution:</strong></p><section class="tib-hiddenbox"><p>We can find the last term - the 25<sup>th</sup> term, <span class="math-tex">\({u_{25}}\)</span> - by using the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term.</p><p><span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\;\;\; \Rightarrow \;\;\;{u_{25}} = 13 + \left( {25 - 1} \right)\left( { - \,4} \right)\;\;\; \Rightarrow \;\;\;{u_{25}} = - \,83\)</span></p><p>Now that we know the last term, we can use the formula <span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\)</span> to find <span class="math-tex">\({S_{25}}\)</span> by substituting for <span class="math-tex">\(n\)</span>, <span class="math-tex">\({{u_1}}\)</span> and <span class="math-tex">\({{u_n}}\)</span>.</p><p><span class="math-tex">\({S_{25}} = \frac{{25}}{2}\left( {13 - 83} \right) = - \,875\)</span> ; hence, the sum of the first 25 terms of the series <span class="math-tex">\(13 + 9 + 5 + 1 - 3 - 7 - \; \cdots \)</span> is <span class="math-tex">\( - \,875\)</span>.</p></section></div><p>The approach used for the solution to Example 4 can be used to develop another formula for the sum of an arithmetic series. As we know from the first sub-section on this page (<strong>1B.1</strong>), the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of an arithmetic sequence is <span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\)</span>. Substituting <span class="math-tex">\({u_1} + \left( {n - 1} \right)d\)</span> for <span class="math-tex">\({u_n}\)</span> in the formula <span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\)</span> gives <span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right) = \frac{n}{2}\left( {{u_1} + {u_1} + \left( {n - 1} \right)d} \right) = \frac{n}{2}\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)</span>.</p><p>Using this version of the formula for the sum of an arithmetic series, the calculation for Example 4 goes as follows:</p><p><span class="math-tex">\({S_{25}} = \frac{{25}}{2}\left( {2 \cdot 13 + \left( {25 - 1} \right)\left( { - \,4} \right)} \right) = \frac{{25}}{2}\left( {26 - 96} \right) = \frac{{25}}{2}\left( { - 70} \right) = - \,875\)</span></p><div class="pinkBg"><h4>Sum of an arithmetic series with <em>n</em> terms (version 2)</h4><p>The sum of <span class="math-tex">\(n\)</span> terms, <span class="math-tex">\({S_n}\)</span>, of an <strong>arithmetic series</strong> with first term <span class="math-tex">\({u_1}\)</span> and common difference <span class="math-tex">\(d\)</span> is given by the following<strong> </strong>formula (in the <a href="/media/ib/mathanalysis/analysis/basics/aa-formula-bklet-v1.2.pdf" target="_blank">Analysis & Approaches formula booklet</a>):</p><p><span class="math-tex">\({S_n} = \frac{n}{2}\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)</span></p></div><p><strong>Example 5</strong></p><p>Let's return to the 30-day job scenario described at the very start of this section. What will be your total payment for the 30 days if you are paid according to Plan A?</p><div class="box"><p><strong>Example 5 solution:</strong></p><section class="tib-hiddenbox"><p>As we indicated earlier, the list of daily payments with Plan A is an arithmetic sequence of 30 terms such that <span class="math-tex">\({u_1} = 1\)</span> and <span class="math-tex">\(d = 3\)</span>. Using the formula for the sum of an arithmetic series, <span class="math-tex">\({S_n} = \frac{n}{2}\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)</span>, sum of the terms in this sequence is: <span class="math-tex">\({S_{30}} = \frac{{30}}{2}\left( {2 \cdot 1 + \left( {30 - 1} \right)3} \right) = 1335\)</span>. Therefore, your total pay for the 30-day job using Plan A is $1335.</p></section></div><hr><h4><a class="anchor" id="sigma" name="sigma"> </a>1B.4 Sigma (<span class="math-tex">\(\Sigma\)</span>) notation</h4><p>Recall from subsection <strong>1A.1</strong> that sigma notation is an efficient way to represent a sum using the symbol <span class="math-tex">\(\sum \)</span> (Greek capital letter <em>sigma</em>). The notation <span class="math-tex">\(\sum\limits_{r = 1}^n {{u_r}} \)</span> represents the sum of <span class="math-tex">\(n\)</span> terms (hence, <span class="math-tex">\(n\)</span> is a constant) where <span class="math-tex">\({{u_r}}\)</span> is the algebraic rule (in terms of the variable <span class="math-tex">\(r\)</span>) for generating each term starting with <span class="math-tex">\(r = 1\)</span>, then <span class="math-tex">\(r = 2\)</span>, and so on, up to <span class="math-tex">\(r = n\)</span>. In other words, what is written below the sigma symbol, <span class="math-tex">\(\Sigma\)</span>, tells us the variable and the starting value for the variable; and the number written above <span class="math-tex">\(\Sigma\)</span> tells us the last value of the variable. The variable will only have integer values. For example, the expression <span class="math-tex">\(\sum\limits_{r = 0}^4 {\left( {2r + 3} \right)} \)</span> tells us to add terms generated by the rule <span class="math-tex">\({u_r} = 2r + 3\)</span> for values of the variable <span class="math-tex">\(r\)</span> starting at <span class="math-tex">\(r = 0\)</span> and stopping at <span class="math-tex">\(r = 4\)</span> (so, there will be 5 terms). Hence, <span class="math-tex">\(\sum\limits_{r = 0}^4 {\left( {2r + 3} \right)} = 3 + 5 + 7 + 9 + 11\)</span>.</p><p>Also, recall from sub-section <strong>1B.2</strong> above that the rule (formula) for terms in an arithmetic sequence will be a <strong>linear function</strong>, and that the terms in an arithmetic sequence display <strong>linear growth</strong> (positive <span class="math-tex">\(d\)</span>) or <strong>linear decay</strong> (negative <span class="math-tex">\(d\)</span>). Thus, <span class="math-tex">\(\sum\limits_{r = 1}^n {{u_r}} \)</span> represents an <strong>arithmetic series</strong> if <span class="math-tex">\({u_r}\)</span> is a linear function which means it is in the form <span class="math-tex">\({u_r} = d \cdot r + c\)</span> where <span class="math-tex">\(d\)</span> is the common difference and <span class="math-tex">\(c\)</span> is also a constant.</p><div class="box"><h4>Examples of series represented with sigma notation</h4><table><tbody><tr><td>(a) <span class="math-tex">\(\sum\limits_{r = 1}^7 r = 1 + 2 + 3 + 4 + 5 + 6 + 7\)</span></td><td>(b) <span class="math-tex">\(\sum\limits_{i = 1}^4 {\left( {3{i^2} - 2i + 1} \right)} = 2 + 9 + 22 + 41\)</span></td></tr><tr><td>(c) <span class="math-tex">\(\sum\limits_{n = 1}^5 {\left( {3n - 8} \right)} = - \,5 - 2 + 1 + 4 + 7 + 10\)</span></td><td>(d) <span class="math-tex">\(\sum\limits_{n = 0}^4 {\frac{1}{{n + 1}}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\)</span></td></tr><tr><td>(e) <span class="math-tex">\({\sum\limits_{r = 0}^\infty {\left( {\frac{1}{2}} \right)} ^r} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \; \cdots \)</span></td><td>(f) <span class="math-tex">\(\sum\limits_{r = 1}^6 {\left( {30 - 8r} \right)} = 22 + 14 + 6 - 2 - 10 - 18\)</span></td></tr></tbody></table><p>Note that the variable (<span class="math-tex">\(r\)</span>, <span class="math-tex">\(n\)</span> or <span class="math-tex">\(i\)</span> are used here) written below the <span class="math-tex">\(\Sigma\)</span> symbol can start at any value but typically the variable starts at 1 or 0. It is convenient for it to start at 1 such as in (a), (b), (c) & (f) because then the value of the variable for a particular term will match the position of the term. For example, in (b) the 3<sup>rd</sup> term has <span class="math-tex">\(i=3\)</span> , so the value of the variable <span class="math-tex">\(i\)</span> also serves as a 'counter' (1 for 1<sup>st</sup> term, 2 for 2<sup>nd</sup> term, etc). The series in (a), (c) & (f) are <strong>arithmetic</strong>. Note that the rule (formula) for each of these is a <strong>linear function</strong>: <span class="math-tex">\({u_r} = r\)</span> in (a); <span class="math-tex">\({u_n} = 3n - 8\)</span> in (c); and <span class="math-tex">\({u_r} = 30 - 8r\)</span> in (f). The series in (b), (d) & (e) are not arithmetic (series in (e) is geometric) so the rule for each of these is not a linear function.</p></div><p><strong>Example 6</strong></p><p>Write an expression using sigma notation to represent each of the following arithmetic series. Use <span class="math-tex">\(r\)</span> for the variable (counter) and start at <span class="math-tex">\(r = 1\)</span>.</p><p>(a) <span class="math-tex">\( - \,11 - 5 + 1 + 7 + 13 + 19 + 25\)</span></p><p>(b) <span class="math-tex">\(29.3 + 26.6 + 23.9 + 21.2 + 18.5\)</span></p><div class="box"><p><strong>Example 6 solution:</strong></p><section class="tib-hiddenbox"><p>(a) Use the formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of an arithmetic sequence, <span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\)</span>, to find the rule (formula) for the series.</p><p><span class="math-tex">\({u_1} = - 11\)</span> and <span class="math-tex">\(d = 6\)</span>; substituting gives <span class="math-tex">\({u_n} = - 11 + \left( {n - 1} \right)6\;\;\; \Rightarrow \;\;\;{u_n} = 6n - 17\)</span></p><p>Therefore, <span class="math-tex">\(\sum\limits_{r = 1}^7 {\left( {6n - 17} \right) = } - \,11 - 5 + 1 + 7 + 13 + 19 + 25\)</span></p><p>(b) Repeating the approach used in (a): </p><p><span class="math-tex">\({u_1} = 29.3\)</span> and <span class="math-tex">\(r = - \,0.7\)</span>; substituting gives <span class="math-tex">\({u_n} = 29.3 + \left( {n - 1} \right)\left( { - \,0.7} \right)\;\;\; \Rightarrow \;\;\;{u_n} = 30 - 0.7n\)</span></p><p>Therefore, <span class="math-tex">\(\sum\limits_{r = 1}^5 {\left( {30 - 0.7n} \right) = } 29.3 + 26.6 + 23.9 + 21.2 + 18.5\)</span></p></section></div><hr><h4><a class="anchor" id="apps" name="apps"> </a>1B.5 Applications of arithemetic sequences & series</h4><p>Arithmetic sequences and series are utilized in a wide range of applications. Examples in the first three sub-sections on this page illustrate some applications. A common application of arithmetic series is <a href="https://www.bankrate.com/glossary/s/simple-interest/" target="_blank">simple interest</a>. Interest can be interpreted two different ways, as either (i) money that is earned during a period of time from money in an account, or (ii) money that is paid (usually in installments) as the cost for borrowing a certain amount of money (the <em>principal</em>). <strong>Simple interest</strong> is calculated from only the (i) initial amount deposited into an account, or (ii) the principal of a loan (amount borrowed); so, the interest for each time interval is the same (constant). In the next section (<strong>1C</strong>) we will see that <strong>compound interest</strong> operates differently in that interest earned from an account, or interest paid for a loan, will not be the same for each time interval.</p><p><strong>Example 7</strong></p><p>If you deposited £1000 in an account that earns simple interest at a rate of 2.45% per year (per annum), how much would be in the account after 10 years? Assume that the account has no further deposits and no withdrawals during the 10 years.</p><div class="box"><p><strong>Example 7 solution:</strong></p><section class="tib-hiddenbox"><p>The simple interest earned each year is a constant which is 2.45% of 1000. Hence, the amount of simple interest added to the account each year is equal to <span class="math-tex">\(\frac{{2.45}}{{100}}\left( {1000} \right) = 24.5\)</span>.<br>The total amount in the account after 10 years will be the sum of an arithmetic series with 11 terms because the first term will be initial deposit of £1000 and then there are 10 further terms for each of the 10 years where interest is added. Also, <span class="math-tex">\({u_1} = 1000\)</span> and <span class="math-tex">\(d = 24.5\)</span>. We do not know the last term in the series, so we will use the formula <span class="math-tex">\({S_n} = \frac{n}{2}\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)</span> to find the sum of the series. Total amount <span class="math-tex">\( = {S_{11}} = \frac{{11}}{2}\left( {2 \cdot 1000 + \left( {11 - 1} \right)24.5} \right) = 12347.50\)</span><br>Therefore, after 10 years the account will have £12,347.50</p></section></div><hr><h3><a class="anchor" id="summary" name="summary"> </a>1B.6 Summary</h3><ul><li>The explicit <strong>formula for the <span class="math-tex">\(n\)</span><sup>th</sup> term of an arithmetic sequence</strong> is: <span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\)</span></li><li>The explicit <strong>formula </strong>for the <span class="math-tex">\(n\)</span><sup>th</sup> term of an <strong>arithmetic sequence</strong> is a <strong>linear function</strong> whose plot is a <strong>set of discrete points</strong>, <span class="math-tex">\(\left( {n,{u_n}} \right)\)</span>, that all lie on a line with a gradient of <span class="math-tex">\(d\)</span>, the common difference of the sequence.</li><li>The terms in an arithmetic sequence display <strong>linear growth</strong> (positive <span class="math-tex">\(d\)</span>) or <strong>linear decay</strong> (negative <span class="math-tex">\(d\)</span>).</li><li>An arithmetic sequence can be used to <strong>model and predict</strong> further results for a set of numbers that display a pattern that is close to being arithmetic but is not perfectly arithmetic (i.e. the difference between terms is constant).</li><li>There are two versions for the <strong>formula to calculate the sum of an arithmetic series</strong>:<br>(i) <span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\)</span> (ii) <span class="math-tex">\({S_n} = \frac{n}{2}\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)</span></li><li><strong>Sigma (<span class="math-tex">\(\Sigma\)</span>) notation</strong> is a convenient and efficient way to represent arithmetic series. <span class="math-tex">\(\sum\limits_{r = 1}^n {{u_r}} \)</span> represents the sum of <span class="math-tex">\(n\)</span> terms where <span class="math-tex">\({{u_r}}\)</span> is the algebraic rule (a linear function) for generating each term.</li><li>Arithmetic sequences and series have many <strong>applications</strong>. For this course, the most important application to be familiar with are questions involving <strong>simple interest</strong>.</li></ul><hr><table width="100%"><tbody><tr><td>← go to previous section: <a href="../34833/1a-sequences-series-intro-sn.html" target="_self">1A. Sequences & series intro</a></td><td style="text-align: right;"> go to next section: <a href="https://www.thinkib.net/mathanalysis/page/35096/1c-geometric-sequences-series-sn" target="_blank">1C. Geometric sequences & series</a> → </td></tr></tbody></table><script>document.querySelectorAll('.tib-teacher-only').forEach(e => e.remove());</script>
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$('img.ico[src="/img/icons/comments.png"]').each((function(){var t=$(this).attr("title");$(this).removeAttr("title"),$(this).wrap('<a class="tib-popover" href="#" data-content="'+t+'" data-togle="popover" data-placement="top" />')})),$('img.ico[src="../../../img/icons/comments.png"]').each((function(){var t=$(this).attr("title");$(this).removeAttr("title"),$(this).wrap('<a class="tib-popover" href="#" data-content="'+t+'" data-togle="popover" data-placement="top" />')})),$(".tib-popover").popover({html:!0,trigger:"hover",delay:{show:300,hide:300},placement:function(t,e){var o=$(e).position();return o.top>200?"top":o.left<515?"right":o.top<200?"bottom":o.left>515?"left":"top"}}).click((function(t){t.preventDefault()}));var carouselTime=6500;$("div.carousel.slide").carousel({interval:carouselTime}),$(".tib-indicators > img").click((function(){var t=$(this).index(),e;$(this).closest(".carousel.slide").carousel(t)})),setShowResultsListeners($("#container")),addMarksThreads($("#modal-std-write"));
$('a.btn.showhider').click(function(e) {
var showHiderId = $( this ).attr('rel');
if( $('#'+showHiderId).find('iframe').length > 0 ) {
$('#'+showHiderId+' iframe').each(function() {
if ( $(this).attr('src').indexOf('.pdf') > 0 ) {
this.contentWindow.location.reload(true);
}
});
}
});
});
</script>
</body>
</html>