{"id":5250,"date":"2018-05-12T18:24:48","date_gmt":"2018-05-12T18:24:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coreq-collegealgebra\/?post_type=chapter&#038;p=5250"},"modified":"2018-05-21T20:08:25","modified_gmt":"2018-05-21T20:08:25","slug":"series-and-their-notations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/series-and-their-notations\/","title":{"raw":"Series and Their Notations*","rendered":"Series and Their Notations*"},"content":{"raw":"A couple decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned.\r\n<h2>Arithmetic Series<\/h2>\r\n<h3>Using Summation Notation<\/h3>\r\nTo find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a <strong>series<\/strong>. Consider, for example, the following series.\r\n<p style=\"text-align: center;\">[latex]3+7+11+15+19+\\cdots[\/latex]<\/p>\r\nThe <strong>[latex]n\\text{th }[\/latex] partial sum<\/strong> of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation\r\n<p style=\"text-align: center;\">[latex]\\text{ }{S}_{n}\\text{ }[\/latex] represents the partial sum.\r\n[latex]\\begin{array}{l}{S}_{1}=3\\\\ {S}_{2}=3+7=10\\\\ {S}_{3}=3+7+11=21\\\\ {S}_{4}=3+7+11+15=36\\end{array}[\/latex]<\/p>\r\n<strong>Summation notation <\/strong>is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter <strong>sigma<\/strong>, [latex]\\sigma[\/latex], to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term in the series. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term in a series.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03225157\/CNX_Precalc_Figure_11_04_001n2.jpg\" alt=\"Explanation of summation notion as described in the text.\" data-media-type=\"image\/jpg\" \/>\r\n\r\nIf we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{k}=2k[\/latex] for [latex]k=1[\/latex] through [latex]k=5[\/latex]. We can begin by substituting the terms for [latex]k[\/latex] and listing out the terms of this series.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ {a}_{1}=2\\left(1\\right)=2\\end{array}\\hfill \\\\ {a}_{2}=2\\left(2\\right)=4\\hfill \\\\ {a}_{3}=2\\left(3\\right)=6\\hfill \\\\ {a}_{4}=2\\left(4\\right)=8\\hfill \\\\ {a}_{5}=2\\left(5\\right)=10\\hfill \\end{array}[\/latex]<\/p>\r\nWe can find the sum of the series by adding the terms:\r\n<p style=\"text-align: center;\">[latex]\\sum _{k=1}^{5}2k=2+4+6+8+10=30[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Summation Notation<\/h3>\r\nThe sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:\r\n<p style=\"text-align: center;\">[latex]\\sum _{k=1}^{n}{a}_{k}[\/latex]<\/p>\r\nThis notation tells us to find the sum of [latex]{a}_{k}[\/latex] from\r\n<p style=\"text-align: center;\">[latex]k=1[\/latex] to [latex]k=n[\/latex].<\/p>\r\n[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h4>Does the lower limit of summation have to be 1?<\/h4>\r\n<em>No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given summation notation for a series, evaluate the value.<\/h3>\r\n<ol>\r\n \t<li>Identify the lower limit of summation.<\/li>\r\n \t<li>Identify the upper limit of summation.<\/li>\r\n \t<li>Substitute each value of [latex]k[\/latex] from the lower limit to the upper limit into the formula.<\/li>\r\n \t<li>Add to find the sum.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Summation Notation<\/h3>\r\nEvaluate [latex]\\sum _{k=3}^{7}{k}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"991305\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"991305\"]\r\n\r\nAccording to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{k}^{2}[\/latex] from [latex]k=3[\/latex] to [latex]k=7[\/latex]. We find the terms of the series by substituting [latex]k=3\\text{,}4\\text{,}5\\text{,}6[\/latex], and [latex]7[\/latex] into the function [latex]{k}^{2}[\/latex]. We add the terms to find the sum.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\sum _{k=3}^{7}{k}^{2}\\hfill &amp; ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill &amp; =9+16+25+36+49\\hfill \\\\ \\hfill &amp; =135\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\sum _{k=2}^{5}\\left(3k - 1\\right)[\/latex].\r\n\r\n[reveal-answer q=\"788016\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"788016\"]38[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5866&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Arithmetic Series<\/h3>\r\nJust as we studied special types of sequences, we will look at special types of series. Recall that an <strong>arithmetic sequence<\/strong> is a sequence in which the difference between any two consecutive terms is the <strong>common difference<\/strong>, [latex]d[\/latex]. The sum of the terms of an arithmetic sequence is called an <strong>arithmetic series<\/strong>. We can write the sum of the first [latex]n[\/latex] terms of an arithmetic series as:\r\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}[\/latex].<\/p>\r\nWe can also reverse the order of the terms and write the sum as\r\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}[\/latex].<\/p>\r\nIf we add these two expressions for the sum of the first [latex]n[\/latex] terms of an arithmetic series, we can derive a formula for the sum of the first [latex]n[\/latex] terms of any arithmetic series.\r\n<p style=\"text-align: center;\">[latex]\\frac{\\begin{array}{l}{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}\\hfill \\\\ +{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}\\hfill \\end{array}}{2{S}_{n}=\\left({a}_{1}+{a}_{n}\\right)+\\left({a}_{1}+{a}_{n}\\right)+...+\\left({a}_{1}+{a}_{n}\\right)}[\/latex]<\/p>\r\nBecause there are [latex]n[\/latex] terms in the series, we can simplify this sum to\r\n<p style=\"text-align: center;\">[latex]2{S}_{n}=n\\left({a}_{1}+{a}_{n}\\right)[\/latex].<\/p>\r\nWe divide by 2 to find the formula for the sum of the first [latex]n[\/latex] terms of an arithmetic series.\r\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">This is generally referred to as the <strong>Partial Sum<\/strong> of the series.<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for the Partial Sum of\u00a0an Arithmetic Series<\/h3>\r\nAn <strong>arithmetic series<\/strong> is the sum of the terms of an arithmetic sequence. The formula for the partial sum of an arithmetic sequence is\r\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given terms of an arithmetic series, find the partial sum<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]{a}_{1}[\/latex] and [latex]{a}_{n}[\/latex].<\/li>\r\n \t<li>Determine [latex]n[\/latex].<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1}\\text{, }{a}_{n}[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the partial sum\u00a0of an Arithmetic Series<\/h3>\r\nFind the partial sum of each arithmetic series.\r\n<ol>\r\n \t<li>[latex]\\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[\/latex]<\/li>\r\n \t<li>[latex]\\text{20 + 15 + 10 +\\ldots + -50}[\/latex]<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{12}3k - 8[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"470866\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"470866\"]\r\n<ol>\r\n \t<li>We are given [latex]{a}_{1}=5[\/latex] and [latex]{a}_{n}=32[\/latex].Count the number of terms in the sequence to find [latex]n=10[\/latex].Substitute values for [latex]{a}_{1},{a}_{n}\\text{\\hspace{0.17em},}[\/latex] and [latex]n[\/latex] into the formula and simplify.\r\n[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\hfill \\end{array}\\hfill \\\\ {S}_{10}=\\frac{10\\left(5+32\\right)}{2}=185\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>We are given [latex]{a}_{1}=20[\/latex] and [latex]{a}_{n}=-50[\/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[\/latex].\r\n[latex]\\begin{array}{l}{a}_{n}={a}_{1}+\\left(n - 1\\right)d\\hfill \\\\ -50=20+\\left(n - 1\\right)\\left(-5\\right)\\hfill \\\\ -70=\\left(n - 1\\right)\\left(-5\\right)\\hfill \\\\ 14=n - 1\\hfill \\\\ 15=n\\hfill \\end{array}[\/latex]\r\nSubstitute values for [latex]{a}_{1},{a}_{n}\\text{,}n[\/latex] into the formula and simplify.[latex]\\begin{array}{l}\\begin{array}{l}\\\\ {S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\end{array}\\hfill \\\\ {S}_{15}=\\frac{15\\left(20 - 50\\right)}{2}=-225\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>To find [latex]{a}_{1}[\/latex], substitute [latex]k=1[\/latex] into the given explicit formula.\r\n[latex]\\begin{array}{l}{a}_{k}=3k - 8\\hfill \\\\ \\text{ }{a}_{1}=3\\left(1\\right)-8=-5\\hfill \\end{array}[\/latex]\r\nWe are given that [latex]n=12[\/latex]. To find [latex]{a}_{12}[\/latex], substitute [latex]k=12[\/latex] into the given explicit formula.\r\n[latex]\\begin{array}{l}\\text{ }{a}_{k}=3k - 8\\hfill \\\\ {a}_{12}=3\\left(12\\right)-8=28\\hfill \\end{array}[\/latex]\r\nSubstitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.\r\n[latex]\\begin{array}{l}\\text{ }{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\hfill \\\\ {S}_{12}=\\frac{12\\left(-5+28\\right)}{2}=138\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the formula to find the partial sum of each arithmetic series.\r\n\r\n[latex]\\text{1}\\text{.4 + 1}\\text{.6 + 1}\\text{.8 + 2}\\text{.0 + 2}\\text{.2 + 2}\\text{.4 + 2}\\text{.6 + 2}\\text{.8 + 3}\\text{.0 + 3}\\text{.2 + 3}\\text{.4}[\/latex]\r\n\r\n[reveal-answer q=\"649728\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"649728\"][latex]\\text{26}\\text{.4}[\/latex][\/hidden-answer]\r\n\r\n[latex]\\text{13 + 21 + 29 + }\\dots \\text{+ 69}[\/latex]\r\n\r\n[reveal-answer q=\"617640\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"617640\"][latex]\\text{328}[\/latex][\/hidden-answer]\r\n\r\n[latex]\\sum _{k=1}^{10}5 - 6k[\/latex]\r\n\r\n[reveal-answer q=\"794771\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"794771\"][latex]\\text{-280}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=128790&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=128791&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Application Problems with Arithmetic Series<\/h3>\r\nOn the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?\r\n\r\n[reveal-answer q=\"455757\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"455757\"]\r\n\r\nThis problem can be modeled by an arithmetic series with [latex]{a}_{1}=\\frac{1}{2}[\/latex] and [latex]d=\\frac{1}{4}[\/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[\/latex], and we are looking for [latex]{S}_{8}[\/latex]. To find [latex]{a}_{8}[\/latex], we can use the explicit formula for an arithmetic sequence.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ {a}_{n}={a}_{1}+d\\left(n - 1\\right)\\end{array}\\hfill \\\\ {a}_{8}=\\frac{1}{2}+\\frac{1}{4}\\left(8 - 1\\right)=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/p>\r\nWe can now use the formula for arithmetic series.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} {S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\hfill \\\\ \\text{ }{S}_{8}=\\frac{8\\left(\\frac{1}{2}+\\frac{9}{4}\\right)}{2}=11\\hfill \\end{array}[\/latex]<\/p>\r\nShe will have walked a total of 11 miles.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?\r\n\r\n[reveal-answer q=\"454197\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"454197\"]$2,025[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5865&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Geometric Series<\/h2>\r\nJust as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a <strong>geometric series<\/strong>. Recall that a <strong>geometric sequence<\/strong> is a sequence in which the ratio of any two consecutive terms is the <strong>common ratio<\/strong>, [latex]r[\/latex]. We can write the sum of the first [latex]n[\/latex] terms of a geometric series as\r\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}[\/latex].<\/p>\r\nJust as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first [latex]n[\/latex] terms of a geometric series. We will begin by multiplying both sides of the equation by [latex]r[\/latex].\r\n<p style=\"text-align: center;\">[latex]r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}[\/latex]<\/p>\r\nNext, we subtract this equation from the original equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\frac{\\begin{array}{l}\\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\\hfill \\\\ -r{S}_{n}=-\\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\\right)\\hfill \\end{array}}{\\left(1-r\\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\\end{array}[\/latex]<\/p>\r\nNotice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for [latex]{S}_{n}[\/latex], divide both sides by\r\n<p style=\"text-align: center;\">[latex]\\left(1-r\\right)[\/latex].\r\n[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for the Sum of the First <em>n<\/em> Terms of a Geometric Series<\/h3>\r\nA <strong>geometric series<\/strong> is the sum of the terms in a geometric sequence. The formula for the sum of the first [latex]n[\/latex] terms of a geometric sequence is represented as\r\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a geometric series, find the sum of the first <em>n<\/em> terms.<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]{a}_{1},r,\\text{and}n[\/latex].<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the First <em>n<\/em> Terms of a Geometric Series<\/h3>\r\nUse the formula to find the indicated partial sum of each geometric series.\r\n<ol>\r\n \t<li>[latex]{S}_{11}[\/latex] for the series [latex]\\text{ 8 + -4 + 2 + }\\dots [\/latex]<\/li>\r\n \t<li>[latex]\\underset{6}{\\overset{k=1}{{\\sum }^{\\text{ }}}}3\\cdot {2}^{k}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"618333\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"618333\"]\r\n<ol>\r\n \t<li>[latex]{a}_{1}=8[\/latex], and we are given that [latex]n=11[\/latex].We can find [latex]r[\/latex] by dividing the second term of the series by the first.\r\n[latex]r=\\frac{-4}{8}=-\\frac{1}{2}[\/latex]\r\nSubstitute values for [latex]{a}_{1}, r, \\text{and} n[\/latex] into the formula and simplify.\r\n[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{11}=\\frac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336\\hfill \\end{array}[\/latex]<\/li>\r\n \t<li>Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\r\n[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]\r\nWe can see from the given explicit formula that [latex]r=2[\/latex]. The upper limit of summation is 6, so [latex]n=6[\/latex].Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.\r\n[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse the formula to find the indicated partial sum of each geometric series.\r\n[latex]{S}_{20}[\/latex] for the series [latex]\\text{ 1,000 + 500 + 250 + }\\dots [\/latex]\r\n\r\n[reveal-answer q=\"922435\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"922435\"][latex]\\approx 2,000.00[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19446&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n[latex]\\sum _{k=1}^{8}{3}^{k}[\/latex]\r\n\r\n[reveal-answer q=\"15208\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"15208\"]9,840[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Application Problem with a Geometric Series<\/h3>\r\nAt a new job, an employee\u2019s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.\r\n\r\n[reveal-answer q=\"636578\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"636578\"]\r\nThe problem can be represented by a geometric series with [latex]{a}_{1}=26,750[\/latex]; [latex]n=5[\/latex]; and [latex]r=1.016[\/latex]. Substitute values for [latex]{a}_{1}[\/latex], [latex]r[\/latex], and [latex]n[\/latex] into the formula and simplify to find the total amount earned at the end of 5 years.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{5}=\\frac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03\\hfill \\end{array}[\/latex]<\/p>\r\nHe will have earned a total of $138,099.03 by the end of 5 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nAt a new job, an employee\u2019s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?\r\n\r\n[reveal-answer q=\"890801\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"890801\"]$275,513.31[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23741&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Annuities<\/h3>\r\nAt the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An <strong>annuity<\/strong> is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% <strong>annual interest<\/strong>, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added.\r\n\r\nWe can find the value of the annuity right after the last deposit by using a geometric series with [latex]{a}_{1}=50[\/latex] and [latex]r=100.5%=1.005[\/latex]. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned.\r\n\r\nWe can find the value of the annuity after [latex]n[\/latex] deposits using the formula for the sum of the first [latex]n[\/latex] terms of a geometric series. In 6 years, there are 72 months, so [latex]n=72[\/latex]. We can substitute [latex]{a}_{1}=50, r=1.005, \\text{and} n=72[\/latex] into the formula, and simplify to find the value of the annuity after 6 years.\r\n<p style=\"text-align: center;\">[latex]{S}_{72}=\\frac{50\\left(1-{1.005}^{72}\\right)}{1 - 1.005}\\approx 4\\text{,}320.44[\/latex]<\/p>\r\nAfter the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of [latex]72\\left(50\\right) = $3,600[\/latex]. This means that because of the annuity, the couple earned $720.44 interest in their college fund.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an initial deposit and an interest rate, find the value of an annuity.<\/h3>\r\n<ol>\r\n \t<li>Determine [latex]{a}_{1}[\/latex], the value of the initial deposit.<\/li>\r\n \t<li>Determine [latex]n[\/latex], the number of deposits.<\/li>\r\n \t<li>Determine [latex]r[\/latex].\r\n<ol>\r\n \t<li>Divide the annual interest rate by the number of times per year that interest is compounded.<\/li>\r\n \t<li>Add 1 to this amount to find [latex]r[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1}\\text{,}r,\\text{and}n[\/latex]\r\ninto the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex], the value of the annuity after [latex]n[\/latex] deposits.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Annuity Problem<\/h3>\r\nA deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?\r\n\r\n[reveal-answer q=\"808488\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"808488\"]\r\n\r\nThe value of the initial deposit is $100, so [latex]{a}_{1}=100[\/latex]. A total of 120 monthly deposits are made in the 10 years, so [latex]n=120[\/latex]. To find [latex]r[\/latex], divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.\r\n<p style=\"text-align: center;\">[latex]r=1+\\frac{0.09}{12}=1.0075[\/latex]<\/p>\r\nSubstitute [latex]{a}_{1}=100\\text{,}r=1.0075\\text{,}\\text{and}n=120[\/latex] into the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, and simplify to find the value of the annuity.\r\n<p style=\"text-align: center;\">[latex]{S}_{120}=\\frac{100\\left(1-{1.0075}^{120}\\right)}{1 - 1.0075}\\approx 19\\text{,}351.43[\/latex]<\/p>\r\nSo the account has $19,351.43 after the last deposit is made.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nAt the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?\r\n\r\n[reveal-answer q=\"786342\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"786342\"]$92,408.18[\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20277&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Key Equations<\/h2>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>sum of the first [latex]n[\/latex]\r\nterms of an arithmetic series<\/td>\r\n<td>[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>sum of the first [latex]n[\/latex]\r\nterms of a geometric series<\/td>\r\n<td>[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>sum of an infinite geometric series with [latex]-1&lt;r&lt;\\text{ }1[\/latex]<\/td>\r\n<td>[latex]{S}_{n}=\\frac{{a}_{1}}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The sum of the terms in a sequence is called a series.<\/li>\r\n \t<li>A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum.<\/li>\r\n \t<li>The sum of the terms in an arithmetic sequence is called an arithmetic series.<\/li>\r\n \t<li>The sum of the first [latex]n[\/latex] terms of an arithmetic series can be found using a formula.<\/li>\r\n \t<li>The sum of the terms in a geometric sequence is called a geometric series.<\/li>\r\n \t<li>The sum of the first [latex]n[\/latex] terms of a geometric series can be found using a formula.<\/li>\r\n \t<li>The sum of an infinite series exists if the series is geometric with [latex]-1&lt;r&lt;1[\/latex].<\/li>\r\n \t<li>If the sum of an infinite series exists, it can be found using a formula.<\/li>\r\n \t<li>An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<strong>annuity<\/strong> an investment in which the purchaser makes a sequence of periodic, equal payments\r\n\r\n<strong>arithmetic series<\/strong> the sum of the terms in an arithmetic sequence\r\n\r\n<strong>diverge<\/strong> a series is said to diverge if the sum is not a real number\r\n\r\n<strong>geometric series<\/strong> the sum of the terms in a geometric sequence\r\n\r\n<strong>index of summation<\/strong> in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation\r\n\r\n<strong>infinite series<\/strong> the sum of the terms in an infinite sequence\r\n\r\n<strong>lower limit of summation<\/strong> the number used in the explicit formula to find the first term in a series\r\n\r\n<strong>nth partial sum<\/strong> the sum of the first [latex]n[\/latex] terms of a sequence\r\n\r\n<strong>series<\/strong> the sum of the terms in a sequence\r\n\r\n<strong>summation notation<\/strong> a notation for series using the Greek letter sigma; it includes an explicit formula and specifies the first and last terms in the series\r\n\r\n<strong>upper limit of summation<\/strong> the number used in the explicit formula to find the last term in a series","rendered":"<p>A couple decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned.<\/p>\n<h2>Arithmetic Series<\/h2>\n<h3>Using Summation Notation<\/h3>\n<p>To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a <strong>series<\/strong>. Consider, for example, the following series.<\/p>\n<p style=\"text-align: center;\">[latex]3+7+11+15+19+\\cdots[\/latex]<\/p>\n<p>The <strong>[latex]n\\text{th }[\/latex] partial sum<\/strong> of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation<\/p>\n<p style=\"text-align: center;\">[latex]\\text{ }{S}_{n}\\text{ }[\/latex] represents the partial sum.<br \/>\n[latex]\\begin{array}{l}{S}_{1}=3\\\\ {S}_{2}=3+7=10\\\\ {S}_{3}=3+7+11=21\\\\ {S}_{4}=3+7+11+15=36\\end{array}[\/latex]<\/p>\n<p><strong>Summation notation <\/strong>is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter <strong>sigma<\/strong>, [latex]\\sigma[\/latex], to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term in the series. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term in a series.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03225157\/CNX_Precalc_Figure_11_04_001n2.jpg\" alt=\"Explanation of summation notion as described in the text.\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>If we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{k}=2k[\/latex] for [latex]k=1[\/latex] through [latex]k=5[\/latex]. We can begin by substituting the terms for [latex]k[\/latex] and listing out the terms of this series.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ {a}_{1}=2\\left(1\\right)=2\\end{array}\\hfill \\\\ {a}_{2}=2\\left(2\\right)=4\\hfill \\\\ {a}_{3}=2\\left(3\\right)=6\\hfill \\\\ {a}_{4}=2\\left(4\\right)=8\\hfill \\\\ {a}_{5}=2\\left(5\\right)=10\\hfill \\end{array}[\/latex]<\/p>\n<p>We can find the sum of the series by adding the terms:<\/p>\n<p style=\"text-align: center;\">[latex]\\sum _{k=1}^{5}2k=2+4+6+8+10=30[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Summation Notation<\/h3>\n<p>The sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\sum _{k=1}^{n}{a}_{k}[\/latex]<\/p>\n<p>This notation tells us to find the sum of [latex]{a}_{k}[\/latex] from<\/p>\n<p style=\"text-align: center;\">[latex]k=1[\/latex] to [latex]k=n[\/latex].<\/p>\n<p>[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Does the lower limit of summation have to be 1?<\/h4>\n<p><em>No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given summation notation for a series, evaluate the value.<\/h3>\n<ol>\n<li>Identify the lower limit of summation.<\/li>\n<li>Identify the upper limit of summation.<\/li>\n<li>Substitute each value of [latex]k[\/latex] from the lower limit to the upper limit into the formula.<\/li>\n<li>Add to find the sum.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Summation Notation<\/h3>\n<p>Evaluate [latex]\\sum _{k=3}^{7}{k}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q991305\">Solution<\/span><\/p>\n<div id=\"q991305\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{k}^{2}[\/latex] from [latex]k=3[\/latex] to [latex]k=7[\/latex]. We find the terms of the series by substituting [latex]k=3\\text{,}4\\text{,}5\\text{,}6[\/latex], and [latex]7[\/latex] into the function [latex]{k}^{2}[\/latex]. We add the terms to find the sum.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}\\sum _{k=3}^{7}{k}^{2}\\hfill & ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill & =9+16+25+36+49\\hfill \\\\ \\hfill & =135\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\sum _{k=2}^{5}\\left(3k - 1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788016\">Solution<\/span><\/p>\n<div id=\"q788016\" class=\"hidden-answer\" style=\"display: none\">38<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5866&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h3>Arithmetic Series<\/h3>\n<p>Just as we studied special types of sequences, we will look at special types of series. Recall that an <strong>arithmetic sequence<\/strong> is a sequence in which the difference between any two consecutive terms is the <strong>common difference<\/strong>, [latex]d[\/latex]. The sum of the terms of an arithmetic sequence is called an <strong>arithmetic series<\/strong>. We can write the sum of the first [latex]n[\/latex] terms of an arithmetic series as:<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}[\/latex].<\/p>\n<p>We can also reverse the order of the terms and write the sum as<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}[\/latex].<\/p>\n<p>If we add these two expressions for the sum of the first [latex]n[\/latex] terms of an arithmetic series, we can derive a formula for the sum of the first [latex]n[\/latex] terms of any arithmetic series.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\begin{array}{l}{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}\\hfill \\\\ +{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}\\hfill \\end{array}}{2{S}_{n}=\\left({a}_{1}+{a}_{n}\\right)+\\left({a}_{1}+{a}_{n}\\right)+...+\\left({a}_{1}+{a}_{n}\\right)}[\/latex]<\/p>\n<p>Because there are [latex]n[\/latex] terms in the series, we can simplify this sum to<\/p>\n<p style=\"text-align: center;\">[latex]2{S}_{n}=n\\left({a}_{1}+{a}_{n}\\right)[\/latex].<\/p>\n<p>We divide by 2 to find the formula for the sum of the first [latex]n[\/latex] terms of an arithmetic series.<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/p>\n<p style=\"text-align: left;\">This is generally referred to as the <strong>Partial Sum<\/strong> of the series.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for the Partial Sum of\u00a0an Arithmetic Series<\/h3>\n<p>An <strong>arithmetic series<\/strong> is the sum of the terms of an arithmetic sequence. The formula for the partial sum of an arithmetic sequence is<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given terms of an arithmetic series, find the partial sum<\/h3>\n<ol>\n<li>Identify [latex]{a}_{1}[\/latex] and [latex]{a}_{n}[\/latex].<\/li>\n<li>Determine [latex]n[\/latex].<\/li>\n<li>Substitute values for [latex]{a}_{1}\\text{, }{a}_{n}[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the partial sum\u00a0of an Arithmetic Series<\/h3>\n<p>Find the partial sum of each arithmetic series.<\/p>\n<ol>\n<li>[latex]\\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[\/latex]<\/li>\n<li>[latex]\\text{20 + 15 + 10 +\\ldots + -50}[\/latex]<\/li>\n<li>[latex]\\sum _{k=1}^{12}3k - 8[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q470866\">Solution<\/span><\/p>\n<div id=\"q470866\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>We are given [latex]{a}_{1}=5[\/latex] and [latex]{a}_{n}=32[\/latex].Count the number of terms in the sequence to find [latex]n=10[\/latex].Substitute values for [latex]{a}_{1},{a}_{n}\\text{\\hspace{0.17em},}[\/latex] and [latex]n[\/latex] into the formula and simplify.<br \/>\n[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ {S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\hfill \\end{array}\\hfill \\\\ {S}_{10}=\\frac{10\\left(5+32\\right)}{2}=185\\hfill \\end{array}[\/latex]<\/li>\n<li>We are given [latex]{a}_{1}=20[\/latex] and [latex]{a}_{n}=-50[\/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[\/latex].<br \/>\n[latex]\\begin{array}{l}{a}_{n}={a}_{1}+\\left(n - 1\\right)d\\hfill \\\\ -50=20+\\left(n - 1\\right)\\left(-5\\right)\\hfill \\\\ -70=\\left(n - 1\\right)\\left(-5\\right)\\hfill \\\\ 14=n - 1\\hfill \\\\ 15=n\\hfill \\end{array}[\/latex]<br \/>\nSubstitute values for [latex]{a}_{1},{a}_{n}\\text{,}n[\/latex] into the formula and simplify.[latex]\\begin{array}{l}\\begin{array}{l}\\\\ {S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\end{array}\\hfill \\\\ {S}_{15}=\\frac{15\\left(20 - 50\\right)}{2}=-225\\hfill \\end{array}[\/latex]<\/li>\n<li>To find [latex]{a}_{1}[\/latex], substitute [latex]k=1[\/latex] into the given explicit formula.<br \/>\n[latex]\\begin{array}{l}{a}_{k}=3k - 8\\hfill \\\\ \\text{ }{a}_{1}=3\\left(1\\right)-8=-5\\hfill \\end{array}[\/latex]<br \/>\nWe are given that [latex]n=12[\/latex]. To find [latex]{a}_{12}[\/latex], substitute [latex]k=12[\/latex] into the given explicit formula.<br \/>\n[latex]\\begin{array}{l}\\text{ }{a}_{k}=3k - 8\\hfill \\\\ {a}_{12}=3\\left(12\\right)-8=28\\hfill \\end{array}[\/latex]<br \/>\nSubstitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.<br \/>\n[latex]\\begin{array}{l}\\text{ }{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\hfill \\\\ {S}_{12}=\\frac{12\\left(-5+28\\right)}{2}=138\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the formula to find the partial sum of each arithmetic series.<\/p>\n<p>[latex]\\text{1}\\text{.4 + 1}\\text{.6 + 1}\\text{.8 + 2}\\text{.0 + 2}\\text{.2 + 2}\\text{.4 + 2}\\text{.6 + 2}\\text{.8 + 3}\\text{.0 + 3}\\text{.2 + 3}\\text{.4}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q649728\">Solution<\/span><\/p>\n<div id=\"q649728\" class=\"hidden-answer\" style=\"display: none\">[latex]\\text{26}\\text{.4}[\/latex]<\/div>\n<\/div>\n<p>[latex]\\text{13 + 21 + 29 + }\\dots \\text{+ 69}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q617640\">Solution<\/span><\/p>\n<div id=\"q617640\" class=\"hidden-answer\" style=\"display: none\">[latex]\\text{328}[\/latex]<\/div>\n<\/div>\n<p>[latex]\\sum _{k=1}^{10}5 - 6k[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q794771\">Solution<\/span><\/p>\n<div id=\"q794771\" class=\"hidden-answer\" style=\"display: none\">[latex]\\text{-280}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=128790&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=128791&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Application Problems with Arithmetic Series<\/h3>\n<p>On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455757\">Solution<\/span><\/p>\n<div id=\"q455757\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem can be modeled by an arithmetic series with [latex]{a}_{1}=\\frac{1}{2}[\/latex] and [latex]d=\\frac{1}{4}[\/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[\/latex], and we are looking for [latex]{S}_{8}[\/latex]. To find [latex]{a}_{8}[\/latex], we can use the explicit formula for an arithmetic sequence.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ {a}_{n}={a}_{1}+d\\left(n - 1\\right)\\end{array}\\hfill \\\\ {a}_{8}=\\frac{1}{2}+\\frac{1}{4}\\left(8 - 1\\right)=\\frac{9}{4}\\hfill \\end{array}[\/latex]<\/p>\n<p>We can now use the formula for arithmetic series.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l} {S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}\\hfill \\\\ \\text{ }{S}_{8}=\\frac{8\\left(\\frac{1}{2}+\\frac{9}{4}\\right)}{2}=11\\hfill \\end{array}[\/latex]<\/p>\n<p>She will have walked a total of 11 miles.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q454197\">Solution<\/span><\/p>\n<div id=\"q454197\" class=\"hidden-answer\" style=\"display: none\">$2,025<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5865&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Geometric Series<\/h2>\n<p>Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a <strong>geometric series<\/strong>. Recall that a <strong>geometric sequence<\/strong> is a sequence in which the ratio of any two consecutive terms is the <strong>common ratio<\/strong>, [latex]r[\/latex]. We can write the sum of the first [latex]n[\/latex] terms of a geometric series as<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}[\/latex].<\/p>\n<p>Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first [latex]n[\/latex] terms of a geometric series. We will begin by multiplying both sides of the equation by [latex]r[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}[\/latex]<\/p>\n<p>Next, we subtract this equation from the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\frac{\\begin{array}{l}\\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\\hfill \\\\ -r{S}_{n}=-\\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\\right)\\hfill \\end{array}}{\\left(1-r\\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\\end{array}[\/latex]<\/p>\n<p>Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for [latex]{S}_{n}[\/latex], divide both sides by<\/p>\n<p style=\"text-align: center;\">[latex]\\left(1-r\\right)[\/latex].<br \/>\n[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for the Sum of the First <em>n<\/em> Terms of a Geometric Series<\/h3>\n<p>A <strong>geometric series<\/strong> is the sum of the terms in a geometric sequence. The formula for the sum of the first [latex]n[\/latex] terms of a geometric sequence is represented as<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a geometric series, find the sum of the first <em>n<\/em> terms.<\/h3>\n<ol>\n<li>Identify [latex]{a}_{1},r,\\text{and}n[\/latex].<\/li>\n<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the First <em>n<\/em> Terms of a Geometric Series<\/h3>\n<p>Use the formula to find the indicated partial sum of each geometric series.<\/p>\n<ol>\n<li>[latex]{S}_{11}[\/latex] for the series [latex]\\text{ 8 + -4 + 2 + }\\dots[\/latex]<\/li>\n<li>[latex]\\underset{6}{\\overset{k=1}{{\\sum }^{\\text{ }}}}3\\cdot {2}^{k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q618333\">Solution<\/span><\/p>\n<div id=\"q618333\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{a}_{1}=8[\/latex], and we are given that [latex]n=11[\/latex].We can find [latex]r[\/latex] by dividing the second term of the series by the first.<br \/>\n[latex]r=\\frac{-4}{8}=-\\frac{1}{2}[\/latex]<br \/>\nSubstitute values for [latex]{a}_{1}, r, \\text{and} n[\/latex] into the formula and simplify.<br \/>\n[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{11}=\\frac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336\\hfill \\end{array}[\/latex]<\/li>\n<li>Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.<br \/>\n[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]<br \/>\nWe can see from the given explicit formula that [latex]r=2[\/latex]. The upper limit of summation is 6, so [latex]n=6[\/latex].Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.<br \/>\n[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use the formula to find the indicated partial sum of each geometric series.<br \/>\n[latex]{S}_{20}[\/latex] for the series [latex]\\text{ 1,000 + 500 + 250 + }\\dots[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q922435\">Solution<\/span><\/p>\n<div id=\"q922435\" class=\"hidden-answer\" style=\"display: none\">[latex]\\approx 2,000.00[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19446&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><br \/>\n[latex]\\sum _{k=1}^{8}{3}^{k}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q15208\">Solution<\/span><\/p>\n<div id=\"q15208\" class=\"hidden-answer\" style=\"display: none\">9,840<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Application Problem with a Geometric Series<\/h3>\n<p>At a new job, an employee\u2019s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q636578\">Solution<\/span><\/p>\n<div id=\"q636578\" class=\"hidden-answer\" style=\"display: none\">\nThe problem can be represented by a geometric series with [latex]{a}_{1}=26,750[\/latex]; [latex]n=5[\/latex]; and [latex]r=1.016[\/latex]. Substitute values for [latex]{a}_{1}[\/latex], [latex]r[\/latex], and [latex]n[\/latex] into the formula and simplify to find the total amount earned at the end of 5 years.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{5}=\\frac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03\\hfill \\end{array}[\/latex]<\/p>\n<p>He will have earned a total of $138,099.03 by the end of 5 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>At a new job, an employee\u2019s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q890801\">Solution<\/span><\/p>\n<div id=\"q890801\" class=\"hidden-answer\" style=\"display: none\">$275,513.31<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23741&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h3>Annuities<\/h3>\n<p>At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An <strong>annuity<\/strong> is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% <strong>annual interest<\/strong>, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added.<\/p>\n<p>We can find the value of the annuity right after the last deposit by using a geometric series with [latex]{a}_{1}=50[\/latex] and [latex]r=100.5%=1.005[\/latex]. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned.<\/p>\n<p>We can find the value of the annuity after [latex]n[\/latex] deposits using the formula for the sum of the first [latex]n[\/latex] terms of a geometric series. In 6 years, there are 72 months, so [latex]n=72[\/latex]. We can substitute [latex]{a}_{1}=50, r=1.005, \\text{and} n=72[\/latex] into the formula, and simplify to find the value of the annuity after 6 years.<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{72}=\\frac{50\\left(1-{1.005}^{72}\\right)}{1 - 1.005}\\approx 4\\text{,}320.44[\/latex]<\/p>\n<p>After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of [latex]72\\left(50\\right) = $3,600[\/latex]. This means that because of the annuity, the couple earned $720.44 interest in their college fund.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an initial deposit and an interest rate, find the value of an annuity.<\/h3>\n<ol>\n<li>Determine [latex]{a}_{1}[\/latex], the value of the initial deposit.<\/li>\n<li>Determine [latex]n[\/latex], the number of deposits.<\/li>\n<li>Determine [latex]r[\/latex].\n<ol>\n<li>Divide the annual interest rate by the number of times per year that interest is compounded.<\/li>\n<li>Add 1 to this amount to find [latex]r[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Substitute values for [latex]{a}_{1}\\text{,}r,\\text{and}n[\/latex]<br \/>\ninto the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex], the value of the annuity after [latex]n[\/latex] deposits.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Annuity Problem<\/h3>\n<p>A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q808488\">Solution<\/span><\/p>\n<div id=\"q808488\" class=\"hidden-answer\" style=\"display: none\">\n<p>The value of the initial deposit is $100, so [latex]{a}_{1}=100[\/latex]. A total of 120 monthly deposits are made in the 10 years, so [latex]n=120[\/latex]. To find [latex]r[\/latex], divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.<\/p>\n<p style=\"text-align: center;\">[latex]r=1+\\frac{0.09}{12}=1.0075[\/latex]<\/p>\n<p>Substitute [latex]{a}_{1}=100\\text{,}r=1.0075\\text{,}\\text{and}n=120[\/latex] into the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, and simplify to find the value of the annuity.<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{120}=\\frac{100\\left(1-{1.0075}^{120}\\right)}{1 - 1.0075}\\approx 19\\text{,}351.43[\/latex]<\/p>\n<p>So the account has $19,351.43 after the last deposit is made.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786342\">Solution<\/span><\/p>\n<div id=\"q786342\" class=\"hidden-answer\" style=\"display: none\">$92,408.18<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20277&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Key Equations<\/h2>\n<table>\n<tbody>\n<tr>\n<td>sum of the first [latex]n[\/latex]<br \/>\nterms of an arithmetic series<\/td>\n<td>[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>sum of the first [latex]n[\/latex]<br \/>\nterms of a geometric series<\/td>\n<td>[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>sum of an infinite geometric series with [latex]-1<r<\\text{ }1[\/latex]<\/td>\n<td>[latex]{S}_{n}=\\frac{{a}_{1}}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The sum of the terms in a sequence is called a series.<\/li>\n<li>A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum.<\/li>\n<li>The sum of the terms in an arithmetic sequence is called an arithmetic series.<\/li>\n<li>The sum of the first [latex]n[\/latex] terms of an arithmetic series can be found using a formula.<\/li>\n<li>The sum of the terms in a geometric sequence is called a geometric series.<\/li>\n<li>The sum of the first [latex]n[\/latex] terms of a geometric series can be found using a formula.<\/li>\n<li>The sum of an infinite series exists if the series is geometric with [latex]-1<r<1[\/latex].<\/li>\n<li>If the sum of an infinite series exists, it can be found using a formula.<\/li>\n<li>An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<p><strong>annuity<\/strong> an investment in which the purchaser makes a sequence of periodic, equal payments<\/p>\n<p><strong>arithmetic series<\/strong> the sum of the terms in an arithmetic sequence<\/p>\n<p><strong>diverge<\/strong> a series is said to diverge if the sum is not a real number<\/p>\n<p><strong>geometric series<\/strong> the sum of the terms in a geometric sequence<\/p>\n<p><strong>index of summation<\/strong> in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation<\/p>\n<p><strong>infinite series<\/strong> the sum of the terms in an infinite sequence<\/p>\n<p><strong>lower limit of summation<\/strong> the number used in the explicit formula to find the first term in a series<\/p>\n<p><strong>nth partial sum<\/strong> the sum of the first [latex]n[\/latex] terms of a sequence<\/p>\n<p><strong>series<\/strong> the sum of the terms in a sequence<\/p>\n<p><strong>summation notation<\/strong> a notation for series using the Greek letter sigma; it includes an explicit formula and specifies the first and last terms in the series<\/p>\n<p><strong>upper limit of summation<\/strong> the number used in the explicit formula to find the last term in a series<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5250\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 5865, 5867. <strong>Authored by<\/strong>: WebWork- Rochester. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>:  IMathAS Community License CC-BY +GPL<\/li><li>Question ID 128790,128791. <strong>Authored by<\/strong>: Day, Alyson. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 20277. <strong>Authored by<\/strong>: Kissel, Kris. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 23741. <strong>Authored by<\/strong>: Shahbazian, Roy, mb McClure, Caren. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":160,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 5865, 5867\",\"author\":\"WebWork- Rochester\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\" IMathAS Community License CC-BY +GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 128790,128791\",\"author\":\"Day, Alyson\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 20277\",\"author\":\"Kissel, Kris\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 23741\",\"author\":\"Shahbazian, Roy, mb McClure, Caren\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5250","chapter","type-chapter","status-publish","hentry"],"part":473,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5250","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/users\/160"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5250\/revisions"}],"predecessor-version":[{"id":5251,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5250\/revisions\/5251"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/473"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5250\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/media?parent=5250"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5250"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5250"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/license?post=5250"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}