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<div id="main-column" class="span9"><article id="sequences-series" style="margin-top: 16px;"><h1 class="section-title">sequences & series</h1><div class="row-fluid"><section class="span12" id="main-content"><!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/REC-html40/loose.dtd">
<html><body><tr><td><div class="tib-quiz"><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>Find the sum of all the integers from 1000 to 2000 (inclusive) which are not multiples of 7.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>Answer: 1,287,286</p></div><div class="q-solution" style="display: none"><p>sum of integers from 1000 to 2000 <u>not</u> multiples of 7 is the sum of integergs from 1000 to 2000 minus the multiples of seven from1000 to 2000.</p><p>find # of integers from 1000 to 2000: <span class="math-tex">\(2000 = 1000 + \left( {n - 1} \right) \cdot 1\;\;\;\; \Rightarrow \;\;\;\;1000 = n - 1\;\;\;\; \Rightarrow \;\;\;\;n = 1001\)</span></p><p><span class="math-tex">\({S_{1001}} = \frac{{1001}}{2}\left( {1000 + 2000} \right) = 1\;501\;500\)</span></p><p>find # of multiples of 7 from 1000 to 2000:</p><p><span class="math-tex">\(\frac{{1000}}{7} \approx 142.86\;\;\;\; \Rightarrow \;\;\;\;7 \cdot 143 = 1001\)</span> <span class="math-tex">\(\frac{{2000}}{7} \approx 285.71\;\;\;\; \Rightarrow \;\;\;\;7 \cdot 285 = 1995\)</span></p><p><span class="math-tex">\(1995 = 1001 + \left( {n - 1} \right) \cdot 7\;\;\; \Rightarrow \;\;\;994 = 7n - 7\;\;\; \Rightarrow \;\;\;7n = 1001\;\;\; \Rightarrow \;\;\;n = 143\)</span></p><p><span class="math-tex">\({S_{143}} = \frac{{143}}{2}\left( {1001 + 1995} \right) = 214\;214\)</span></p><p>sum of integers 1000 to 2000 minus multiples of 7 from 1000 to 2000 <span class="math-tex">\(=1501500 - 214214=1287286\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>Find the positive constants <span class="math-tex">\(a\)</span> and <span class="math-tex">\(b\)</span> such that <span class="math-tex">\(\frac{1}{4},\;a,\;9\)</span> are in geometric progression and <span class="math-tex">\(\frac{1}{4},\;a,\;9 - b\)</span> are in arithmetic progression.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>Answer: <span class="math-tex">\(a = \frac{3}{2},\;b = \frac{{25}}{4}\)</span></p></div><div class="q-solution" style="display: none"><p><span class="math-tex">\(\frac{a}{{\;\frac{1}{4}}} = \frac{9}{a}\;\;\;\; \Rightarrow \;\;\;\;{a^2} = \frac{9}{4}\;\;\;\; \Rightarrow \;\;\;\;a = \frac{3}{2}\)</span></p><p><span class="math-tex">\(a - \frac{1}{4} = 9 - b - a\;\;\;\; \Rightarrow \;\;\;\;b = \frac{{37}}{4} - 2a\;\;\;\; \Rightarrow \;\;\;\;b = \frac{{37}}{4} - 2\left( {\frac{3}{2}} \right) = \frac{{37}}{4} - \frac{{12}}{4} = \frac{{25}}{4}\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>Find the sum of the following arithmetic series: <span class="tib-mathml"><math><semantics><mrow><mn>3</mn><mtext> </mtext><mo>+</mo><mtext> </mtext><mn>7</mn><mtext> </mtext><mo>+</mo><mn>11</mn><mtext> </mtext><mo>+</mo><mtext> </mtext><mtext> </mtext><mo>⋯</mo><mtext> </mtext><mtext> </mtext><mo>+</mo><mtext> </mtext><mn>399</mn></mrow></semantics></math></span></p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>answer: 20100</p></div><div class="q-solution" style="display: none"><p><span class="math-tex">\({u_1} = 3,\;d = 4\)</span></p><p><span class="math-tex">\({u_n} = {u_1} + (n - 1)d\;\;\; \Rightarrow \;\;\;399 = 3 + (n - 1)4\;\;\; \Rightarrow \;\;\;n = \frac{{396}}{4} + 1 = 100\)</span></p><p><span class="math-tex">\({S_n} = \frac{n}{2}({u_1} + {u_n}) = \frac{{100}}{2}(3 + 399) = 20100\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>In an arithmetic series, the first term is -7 and the sum of the first 20 terms is 620.</p><p>(a) Find the common difference.</p><p>(b) Find the value of the 78<sup>th</sup> term.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>(a) <span class="math-tex">\(d = 4\)</span> (b) <span class="math-tex">\({u_{78}} = 301\)</span></p></div><div class="q-solution" style="display: none"><p>(a) <span class="math-tex">\({S_{20}} = \frac{{20}}{2}\left[ {2\left( { - 7} \right) + \left( {20 - 1} \right)d} \right]\)</span></p><p><span class="math-tex">\(620 = 10\left( { - 14 + 19d} \right)\)</span></p><p><span class="math-tex">\(190d = 760\)</span></p><p><span class="math-tex">\(d = \frac{{760}}{{190}} = 4\)</span></p><p>(b) <span class="math-tex">\({u_{78}} = - 7 + \left( {78 - 1} \right)4\)</span></p><p><span class="math-tex">\({u_{78}} = - 7 + 308 = 301\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>Find the sum of the multiples of 9 between 100 and 500.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>13266</p></div><div class="q-solution" style="display: none"><p><span class="math-tex">\({u_1} = 108,\;{u_n} = 495\)</span>, <span class="math-tex">\(108 = 9 \cdot 12;\;\;\;495 = 9 \cdot 55\)</span></p><p><span class="math-tex">\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\)</span>, need to find <span class="math-tex">\(n\)</span></p><p><span class="math-tex">\({u_n} = {u_1} + \left( {n - 1} \right)d\;\;\; \Rightarrow \;\;\;495 = 108 + \left( {n - 1} \right)9\;\;\; \Rightarrow \;\;\;9n = 396\;\;\; \Rightarrow \;\;\;n = 44\)</span></p><p><span class="math-tex">\({S_{44}} = \frac{{44}}{2}\left( {108 + 495} \right) = 13\,266\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>An infinite geometric series with first term 2 converges to the sum of 3. Find the <span class="math-tex">\(4^{th}\)</span> term in the series.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>Answer: <span class="math-tex">\({u_4} = \frac{2}{{27}}\)</span></p></div><div class="q-solution" style="display: none"><p><span class="math-tex">\({u_1} = 2\)</span> and <span class="math-tex">\({S_\infty } = \frac{{{u_1}}}{{1 - r}} = \frac{2}{{1 - r}} = 3\;\;\;\; \Rightarrow \;\;\;\;1 - r = \frac{2}{3}\;\;\;\; \Rightarrow \;\;\;\;r = \frac{1}{3}\)</span></p><p><span class="math-tex">\({u_n} = {u_1}{r^{n - 1}}\)</span>; <span class="math-tex">\({u_4} = 2{\left( {\frac{1}{3}} \right)^3} = \frac{2}{{27}}\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>The sum to infinity of a geometric series is <span class="math-tex">\(\frac{9}{4}\)</span>. The sum of the first four terms is <span class="math-tex">\(\frac{20}{9}\)</span>. Given that the common ratio is negative, find the value of the first term of the series.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>Answer: <span class="math-tex">\({u_1} = 3\)</span></p></div><div class="q-solution" style="display: none"><p><span class="math-tex">\({S_\infty } = \frac{{{u_1}}}{{1 - r}} = \frac{9}{4}\;\;\;\; \Rightarrow \;\;\;\;{u_1} = \frac{9}{4}\left( {1 - r} \right)\)</span></p><p><span class="math-tex">\({u_1} + {u_1}r + {u_1}{r^2} + {u_1}{r^3} = \frac{{20}}{9}\;\;\; \Rightarrow \;\;\;\frac{9}{4}\left( {1 - r} \right) + \frac{9}{4}\left( {1 - r} \right)r + \frac{9}{4}\left( {1 - r} \right){r^2} + \frac{9}{4}\left( {1 - r} \right){r^3} = \frac{{20}}{9}\)</span></p><p>multiply both sides by <span class="math-tex">\(\frac{4}{9}\)</span>: <span class="math-tex">\(1 - r + \left( {1 - r} \right)r + \left( {1 - r} \right){r^2} + \left( {1 - r} \right){r^3} = \frac{{80}}{{81}}\)</span></p><p><span class="math-tex">\(1 - r + r - {r^2} + {r^2} - {r^3} + {r^3} - {r^4} = \frac{{80}}{{81}}\;\;\;\; \Rightarrow \;\;\;\;1 - {r^4} = \frac{{80}}{{81}}\)</span></p><p><span class="math-tex">\({r^4} = \frac{1}{{81}}\;\;\;\; \Rightarrow \;\;\;\;r = \pm \,\frac{1}{3}\)</span> given that <span class="math-tex">\(r\)</span> is negative; thus <span class="math-tex">\(r = - \frac{1}{3}\)</span></p><p>then <span class="math-tex">\({u_1} = \frac{9}{4}\left( {1 + \frac{1}{3}} \right) = 3\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>Alice and Beatriz are practicing penalty kicks during football training. The probability that Alice successfully makes a penalty kick is 0.7, and the probability that Beatriz is successful is 0.8. Alice and Beatriz decide to play a game where they take turns shooting a penalty kick. It is decided that Alice goes first, and a game is won by the first person to make a penalty kick.</p><p>(a) What is the probability that Alice wins on her 2<sup>nd</sup> kick?</p><p>(b) What is the probability that Beatriz wins on her 3<sup>rd</sup> kick?</p><p>(c) For any single game, what is the probability that Beatriz wins?</p><p>(d) For any single game, what is the probability that Alice wins?</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p>Answers: (a) 0.042 (b) 0.0000864 (c) <span class="math-tex">\(\frac{{12}}{{47}} \approx 0.255\)</span> (d) <span class="math-tex">\(\frac{{35}}{{47}} \approx 0.745\)</span></p></div><div class="q-solution" style="display: none"><p>(a) probability Alice wins on second kick <span class="math-tex">\( = \left( {0.3} \right)\left( {0.2} \right)\left( {0.7} \right) = 0.042\)</span></p><p>(b) probability Beatriz wins on 3<sup>rd</sup> kick<span class="math-tex">\( = \left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.8} \right) = 0.0000864\)</span></p><p>(c) probability that Beatriz wins a game is equal to probability that she wins on 1st kick plus probability that she wins on 2nd kick plus probability that she wins on 3rd kick, etc. </p><p>probability B wins game <span class="math-tex">\( = \left( {0.3} \right)\left( {0.8} \right) + \left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.8} \right) + \left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.8} \right) + \; \cdots \)</span></p><p><span class="math-tex">\( = \left( {0.3} \right)\left( {0.8} \right) + {\left( {0.3} \right)^2}\left( {0.2} \right)\left( {0.8} \right) + {\left( {0.3} \right)^3}{\left( {0.2} \right)^2}\left( {0.8} \right) + \; \cdots \)</span></p><p>this is an infinite geometric series with <span class="math-tex">\({u_1} = \left( {0.3} \right)\left( {0.8} \right) = 0.24\)</span>, and <span class="math-tex">\(r = \left( {0.3} \right)\left( {0.2} \right) = 0.06\)</span>.</p><p>The probability that Beatriz wins a game is the sum of this infinite geometric series.</p><p>Thus, probability Beatriz wins a game is <span class="math-tex">\({S_\infty } = \frac{{0.24}}{{1 - 0.06}} \approx 0.255\)</span> [exact value is 12/47]</p><p>(d) probability that Alice wins a game is the <em>complement</em> of Beatriz’ probability of winning.</p><p>thus, probability that Alice wins a game is <span class="math-tex">\(1 - 0.25531915 \approx 0.745\)</span> [exact value is 35/47]</p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>The numbers represented by <span class="math-tex">\(n - 2,\;n,\;n + 3\)</span> are the first three terms, respectively, of a geometric sequence. Find the value of <span class="math-tex">\(n\)</span> and the sum of the first five terms of the sequence.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p><u>Answer</u>:</p><p><span class="math-tex">\(n = 6,\;{S_5} = \frac{{211}}{4} = 52.75\)</span></p></div><div class="q-solution" style="display: none"><p>In a geometric sequence, the ratio of a term and previous term is constant (i.e. common ratio <span class="math-tex">\(r\)</span>)</p><p> <span class="math-tex">\(r = \frac{n}{{n - 2}} = \frac{{n + 3}}{n}\;\; \Rightarrow \;\;{n^2} = \left( {n - 2} \right)\left( {n + 3} \right)\)</span></p><p><span class="math-tex">\({n^2} = {n^2} + n - 6\;\; \Rightarrow \;\;n - 6 = 0\;\; \Rightarrow \;\;n = 6\)</span></p><p><span class="math-tex">\({u_1} = 6 - 2 = 4\)</span> and <span class="math-tex">\({u_2} = 6\)</span>; hence, <span class="math-tex">\(r = \frac{6}{4} = \frac{3}{2}\)</span></p><p>Thus, <span class="math-tex">\({S_5} = \frac{{4\left( {{{\left( {\frac{3}{2}} \right)}^5} - 1} \right)}}{{\frac{3}{2} - 1}} = \frac{{211}}{4}\)</span></p></div></div><div class="exercise" style="margin: 0 0 20px 0;"><div class="q-question"><a class="btn btn-info btn-mini show-answer pull-right">Answer</a><p>Find the sum of all numbers divisible by five between 99 and 301.</p></div><div class="q-answer" style="display: none"><a class="btn btn-info btn-mini show-solution pull-right">Worked solution</a><p><u>Answer</u>:</p><p>8200</p></div><div class="q-solution" style="display: none"><p><span class="math-tex">\({u_1} = 100\)</span> and the last term <span class="math-tex">\({u_n} = 300\)</span>; find <span class="math-tex">\(n\)</span>, the number of terms in the sequence</p><p><span class="math-tex">\({u_n} = 300 = 100 + \left( {n - 1} \right)5\;\; \Rightarrow \;\;n - 1 = \frac{{200}}{5}\;\;\; \Rightarrow \;\;n = 41\)</span></p><p><span class="math-tex">\({S_{41}} = \frac{{41}}{2}\left( {100 + 300} \right) = 8200\)</span></p></div></div><div class="dedicated-uploader" style="border-radius: 4px; border: 1px solid #ccc;"><div class="header" style="padding: 10px; background-color: #f4f4f4;">Click "Upload" to locate and upload your answer(s) and then submit it</div><form id="dedicated-upload-form" enctype="multipart/form-data" method="post" action="/pages/std/std-file-upload.php" style="padding: 10px;"><input type="hidden" name="ticket" value=""><input type="hidden" name="filename" value=""><input type="hidden" name="typeUpload" value="fileUpload"><label class="form-label btn" for="upload-dedicated-attachment" style="background-color: #f4f4f4;"><i class="fa fa-upload" style="margin-right: 5px;"></i>Upload</label><span class="total-files" style="margin-left: 5px;"></span><input id="upload-dedicated-attachment" type="file" name="upload" data-max="1" style="padding: 10px; visibility: hidden;" class="form-control" multiple=""></form></div><div class="done-task" style=""><a href="#" class="btn btn-success btn-large mark-as-done disabled" id="submit-upload-answer" data-type="written-answer" data-task-id="417720"><i class="fa fa-fw fa-check-square-o"></i> Submit</a><span id="task-completed-badge" title="You have successfully submitted this task" class="badge badge-info" style="display: none"><i class="fa fa-fw fa-check"></i> Submitted</span></div></div></td></tr><script>document.querySelectorAll('.tib-teacher-only').forEach(e => e.remove());</script></body></html>
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