{"id":2497,"date":"2016-11-03T22:58:31","date_gmt":"2016-11-03T22:58:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=2497"},"modified":"2017-04-12T17:03:12","modified_gmt":"2017-04-12T17:03:12","slug":"the-formula-for-a-geometric-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/the-formula-for-a-geometric-series\/","title":{"raw":"Geometric Series","rendered":"Geometric Series"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Write the first n terms of a geometric sequence<\/li>\r\n \t<li>Solve an annuity\u00a0problem using a geometric series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n&nbsp;\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\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:\/\/www.myopenmath.com\/multiembedq.php?id=19446&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/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:\/\/www.myopenmath.com\/multiembedq.php?id=23741&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n<\/div>\r\n<h2>Annuities<\/h2>\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:\/\/www.myopenmath.com\/multiembedq.php?id=20277&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\"><\/iframe>\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Write the first n terms of a geometric sequence<\/li>\n<li>Solve an annuity\u00a0problem using a geometric series<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<p>&nbsp;<br \/>\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<\/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:\/\/www.myopenmath.com\/multiembedq.php?id=19446&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/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:\/\/www.myopenmath.com\/multiembedq.php?id=23741&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\n<\/div>\n<h2>Annuities<\/h2>\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:\/\/www.myopenmath.com\/multiembedq.php?id=20277&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\"><\/iframe>\n<\/div>\n<p>&nbsp;<\/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-2497\">\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>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>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 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 LicenseCC-BY + GPL<\/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":21,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 20277\",\"author\":\"Kissel,Kris\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and 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