{"id":450,"date":"2019-07-15T22:45:10","date_gmt":"2019-07-15T22:45:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/the-formula-for-a-geometric-series\/"},"modified":"2019-07-15T22:45:10","modified_gmt":"2019-07-15T22:45:10","slug":"the-formula-for-a-geometric-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/the-formula-for-a-geometric-series\/","title":{"raw":"Geometric Series","rendered":"Geometric Series"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Write the first n terms of a geometric sequence.<\/li>\n \t<li class=\"li2\"><span class=\"s1\">Determine whether the sum of an in\ufb01nite geometric series exists.<\/span><\/li>\n \t<li>Give the sum of a convergent infinite geometric series.<\/li>\n \t<li class=\"li2\">Solve an annuity&nbsp;problem using a geometric series.<\/li>\n<\/ul>\n<\/div>\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\n<p style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+{a}_{1}r+{a}_{1}{r}^{2}+...+{a}_{1}{r}^{n - 1}[\/latex].<\/p>\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].\n<p style=\"text-align: center;\">[latex]r{S}_{n}={a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+...+{a}_{1}{r}^{n}[\/latex]<\/p>\nNext, we subtract this equation from the original equation.\n<p style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;={a}_{1}+{a}_{1}r+{a}_{1}{r}^{2}+...+{a}_{1}{r}^{n - 1} \\\\ -r{S}_{n}&amp;=-\\left({a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+...+{a}_{1}{r}^{n}\\right) \\\\ \\hline \\left(1-r\\right){S}_{n}&amp;={a}_{1}-{a}_{1}{r}^{n}\\end{align}[\/latex]<\/p>\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], factor [latex]a_1[\/latex] on the right hand side and divide both sides by [latex]\\left(1-r\\right)[\/latex].\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\n\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nThe formulas for finding the sum of the first [latex]n[\/latex] terms of arithmetic and geometric series are handy and should be memorized although it is important to understand how they were derived.\n\n<\/div>\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>\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\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\n\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 \t<li>Identify [latex]{a}_{1},r,\\text{and}n[\/latex].<\/li>\n \t<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n \t<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>\nUse the formula to find the indicated partial sum of each geometric series.\n<ol>\n \t<li>[latex]{S}_{11}[\/latex] for the series [latex] 8 + -4 + 2 + \\dots [\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^6 3\\cdot {2}^{k}[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"618333\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"618333\"]\n<ol>\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.\n[latex]r=\\dfrac{-4}{8}=-\\frac{1}{2}[\/latex]\nSubstitute values for [latex]{a}_{1}, r, \\text{and} n[\/latex] into the formula and simplify.\n[latex]\\begin{align}&amp;{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &amp;{S}_{11}=\\dfrac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336 \\\\ \\text{ } \\end{align}[\/latex]<\/li>\n \t<li>Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\n[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]\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.\n[latex]\\begin{align}\\\\ &amp;{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &amp;{S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378 \\end{align}[\/latex]<\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nUse the formula to find the indicated partial sum of each geometric series.\n[latex]{S}_{20}[\/latex] for the series [latex]1\\text{,}000 + 500 + 250 + \\dots [\/latex]\n\n[reveal-answer q=\"922435\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"922435\"]\n\n[latex]\\approx 2,000.00[\/latex]\n\n[\/hidden-answer]\n\n&nbsp;\n\nUse the formula to determine the sum&nbsp;[latex]\\sum\\limits _{k=1}^{8}{3}^{k}[\/latex]\n\n[reveal-answer q=\"15208\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"15208\"]\n\n9,840\n\n[\/hidden-answer]\n\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19446&amp;theme=oea&amp;iframe_resize_id=mom5[\/embed]\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Application Problem with a Geometric Series<\/h3>\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.\n\n[reveal-answer q=\"636578\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"636578\"]\n\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.\n<p style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{5}&amp;=\\dfrac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03 \\end{align}[\/latex]<\/p>\nHe will have earned a total of $138,099.03 by the end of 5 years.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\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?\n\n[reveal-answer q=\"890801\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"890801\"]\n\n$275,513.31\n\n[\/hidden-answer]\n\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5865&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\n\n<\/div>\n<h2>Using the Formula for the Sum of an Infinite Geometric Series<\/h2>\nThus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first&nbsp;<em>n<\/em> terms. An&nbsp;<strong>infinite series<\/strong>&nbsp;is the sum of the terms of an infinite sequence. An example of an infinite series is [latex]2+4+6+8+\\dots[\/latex].\n\nThis series can also be written in summation notation as [latex] \\sum\\limits _{k=1}^{\\infty} 2k[\/latex],&nbsp;where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series&nbsp;<strong>diverges<\/strong>.\n<h3>Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h3>\nIf the terms of an&nbsp;<span class=\"no-emphasis\">infinite geometric series<\/span>&nbsp;approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:\n<p style=\"text-align: center;\">[latex]1+0.2+0.04+0.008+0.0016+\\dots[\/latex]<\/p>\nThe common ratio is [latex]r=0.2[\/latex]. As&nbsp;<em>n<\/em> gets large, the values of of [latex]r^n[\/latex] get very small and approach 0.&nbsp;Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with [latex]-1&lt;r&lt;1[\/latex]&nbsp;approach 0; the sum of a geometric series is defined when&nbsp;[latex]-1&lt;r&lt;1[\/latex].\n<div class=\"textbox\">\n<h3>DETERMINING WHETHER THE SUM OF AN INFINITE GEOMETRIC SERIES IS DEFINED<\/h3>\nThe sum of an infinite series is defined if the series is geometric and&nbsp;[latex]-1&lt;r&lt;1[\/latex].\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: <strong>Given the first several terms of an infinite series, determine if the sum of the series exists.<\/strong><\/h3>\n<ol>\n \t<li>Find the ratio of the second term to the first term.<\/li>\n \t<li>Find the ratio of the third term to the second term.<\/li>\n \t<li>Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.<\/li>\n \t<li>If a common ratio,&nbsp;<em>r<\/em>, was found in step 3, check to see if&nbsp;[latex]-1&lt;r&lt;1[\/latex].&nbsp;If so, the sum is defined. If not, the sum is not defined.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nRemember to try the examples below before looking at the answers!\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:&nbsp;Determining Whether the Sum of an Infinite Series is Defined<\/h3>\nDetermine whether the sum of each infinite series is defined.\n<ol>\n \t<li>[latex]12+8+4+\\dots[\/latex]<\/li>\n \t<li>[latex]\\dfrac{3}{4}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dots[\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty}{27}\\cdot\\left(\\dfrac{1}{3}\\right)^k[\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty}{5k}[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"250515\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"250515\"]\n<ol>\n \t<li>The ratio of the second term to the first is [latex]\\frac{2}{3}[\/latex], which is not the same as the ratio of the third term to the second, [latex]\\frac{1}{2}[\/latex].&nbsp;The series is not geometric.<\/li>\n \t<li>The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of [latex]\\frac{2}{3}[\/latex]. The sum of the infinite series is defined.<\/li>\n \t<li>The given formula is exponential with a base of [latex]\\frac{1}{3}[\/latex]; the series is geometric with a common ratio of&nbsp;[latex]\\frac{1}{3}[\/latex]. The sum of the infinite series is defined.<\/li>\n \t<li>The given formula is not exponential. The series is arithmetic, not geometric and so cannot yield a finite sum.<\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\nDetermine whether the sum of the infinite series is defined.\n<ol>\n \t<li>[latex]\\dfrac{1}{3}+\\dfrac{1}{2}+\\dfrac{3}{4}+\\dfrac{9}{8}+\\cdots[\/latex]<\/li>\n \t<li>[latex]24+(-12)+6+(-3)+\\dots[\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty} 15\\cdot(-0.3)^k[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"559520\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"559520\"]\n<ol>\n \t<li>The series is geometric, but [latex]r=\\dfrac{3}{2}&gt;1[\/latex]. The sum is not defined.<\/li>\n \t<li>The series is geometric with [latex]r=-\\dfrac{1}{2}[\/latex]. The sum is defined.<\/li>\n \t<li>The series is geometric with [latex]r=-0.3[\/latex]. The sum is defined.<\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<h2>Finding Sums of Infinite Series<\/h2>\n<p id=\"fs-id1165137679221\">When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first <em>n<\/em>&nbsp;terms of a geometric series.<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex]<\/p>\nWe will examine an infinite series with [latex]r=\\frac{1}{2}[\/latex]. What happens to [latex]r^n[\/latex] as&nbsp;<em>n<\/em> increases?\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;{\\left(\\frac{1}{2}\\right)}^{2} = \\frac{1}{4} \\\\&amp;{\\left(\\frac{1}{2}\\right)}^{3} = \\frac{1}{8} \\\\&amp;{\\left(\\frac{1}{2}\\right)}^{4} = \\frac{1}{16} \\end{align}[\/latex]<\/p>\nThe value of [latex]r^n[\/latex] decreases rapidly. What happens for greater values of&nbsp;<em>n<\/em>?\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;{\\left(\\frac{1}{2}\\right)}^{10} = \\frac{1}{1\\text{,}024} \\\\&amp;{\\left(\\frac{1}{2}\\right)}^{20} = \\frac{1}{1\\text{,}048\\text{,}576} \\\\&amp;{\\left(\\frac{1}{2}\\right)}^{30} = \\frac{1}{1\\text{,}073\\text{,}741\\text{,}824} \\end{align}[\/latex]<\/p>\nAs&nbsp;<em>n<\/em> gets large, [latex]r^n[\/latex] gets very small. We say that as&nbsp;<em>n<\/em> increases without bound,&nbsp;[latex]r^n[\/latex] approaches 0. As&nbsp;[latex]r^n[\/latex] approaches 0,&nbsp;[latex]1-r^n[\/latex] approaches 1. When this happens the numerator approaches [latex]a_1[\/latex]. This gives us the formula for the sum of an infinite geometric series.\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nThis is another very handy formula that should be memorized. It is important to understand, though, how it is derived from the formula for the&nbsp;sum of the first&nbsp;[latex]n[\/latex]&nbsp;terms of a geometric series for [latex]n[\/latex] increasing without bound.\n\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: FORMULA FOR THE SUM OF AN INFINITE GEOMETRIC SERIES<\/h3>\nThe formula for the sum of an infinite geometric series with [latex]-1&lt;r&lt;1[\/latex] is:\n<p style=\"text-align: center;\">[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: <strong>Given an infinite geometric series, find its sum.<\/strong><\/h3>\n<ol>\n \t<li>Identify [latex]a_1[\/latex] and&nbsp;<em>r<\/em>.<\/li>\n \t<li>Confirm that [latex]-1&lt;r&lt;1[\/latex].<\/li>\n \t<li>Substitute values for&nbsp;[latex]a_1[\/latex] and&nbsp;<em>r<\/em> into the formula,&nbsp;[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex].<\/li>\n \t<li>Simplify to find&nbsp;<em>S<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:&nbsp;Finding the Sum of an Infinite Geometric Series<\/h3>\nFind the sum, if it exists, for the following:\n<ol>\n \t<li>[latex]10+9+8+7+\\dots[\/latex]<\/li>\n \t<li><span id=\"MJXp-Span-5192\" class=\"MJXp-mo\">[latex]248.6+99.44+39.776+\\dots[\/latex]<\/span><\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty}4\\text{,}374\\cdot\\left(-\\dfrac{1}{3}\\right)^{k-1}[\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty}\\dfrac{1}{9}\\cdot\\left(\\dfrac{4}{3}\\right)^{k}[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"18513\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"18513\"]\n<ol>\n \t<li>There is not a constant ratio; the series is not geometric.<\/li>\n \t<li>There is a constant ratio; the series is geometric. [latex]a_1=248.6[\/latex] and [latex]r=\\dfrac{99.44}{248.6}=0.4[\/latex], so the sum exists. Substitute&nbsp;[latex]a_1=248.6[\/latex] and [latex]r=0.4[\/latex] into the formula and simplify to find the sum.\n[latex]\\begin{align} \\\\ &amp;S=\\frac{a_1}{1-r} \\\\[1.5mm] &amp;S=\\frac{248.6}{1-0.4}=\\frac{1243}{3} \\\\ \\text{ }\\end{align}[\/latex]<\/li>\n \t<li>The formula is exponential, so the series is geometric with [latex]r=-\\frac{1}{3}[\/latex]. Find [latex]a_1[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\n[latex]\\begin{align} \\\\ a_1=4\\text{,}374\\cdot\\left(-\\frac{1}{3}\\right)^{1-1}=4\\text{,}374 \\\\ \\text{ }\\end{align}[\/latex]\nSubstitute [latex]4\\text{,}374[\/latex] and [latex]r=-\\frac{1}{3}[\/latex] into the formula, and simplify to find the sum.\n[latex]\\begin{align}\\\\&amp;S=\\frac{a_1}{1-r} \\\\[1.5mm] &amp;S=\\frac{4\\text{,}374}{1-\\left(-\\frac{1}{3}\\right)}=3\\text{,}280.5 \\\\ \\text{ }\\end{align}[\/latex]<\/li>\n \t<li>The formula is exponential, so the series is geometric, but [latex]r&gt;1[\/latex].&nbsp;The sum does not exist.<\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding an Equivalent Fraction for a Repeating Decimal<\/h3>\nFind the equivalent fraction for the repeating decimal [latex]0.\\overline{3}[\/latex].\n\n[reveal-answer q=\"624358\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"624358\"]\n\nWe notice the repeating decimal [latex]0.\\overline{3}=0.333\\dots[\/latex].&nbsp;so we can rewrite the repeating decimal as a sum of terms.\n<p style=\"text-align: center;\">[latex]0.\\overline{3}=0.3+0.03+0.003+\\dots[\/latex]<\/p>\nLooking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term.\n<p style=\"text-align: center;\">[latex]\\begin{align}0.\\overline{3}&amp;=0.3+0.3\\cdot(0.1)+0.3\\cdot(0.01)+0.3\\cdot(0.001)+\\dots \\\\ &amp;=0.3+0.3\\cdot(0.1)+0.3\\cdot(0.1)^2+0.3\\cdot(0.1)^3+\\dots\\end{align}[\/latex]<\/p>\nNotice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have\n<p style=\"text-align: center;\">[latex]S=\\dfrac{a_1}{1-r} =\\dfrac{0.3}{1-0.1} =\\dfrac{0.3}{0.9} =\\dfrac{1}{3}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\nFind the sum if it exists.\n<ol>\n \t<li>[latex]2+\\dfrac{2}{3}+\\dfrac{2}{9}+\\dots[\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty}{0.76k+1}[\/latex]<\/li>\n \t<li>[latex]\\sum\\limits _{k=1}^{\\infty}\\left(-\\dfrac{3}{8}\\right)^k[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"221023\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"221023\"]\n<ol>\n \t<li>&nbsp;3<\/li>\n \t<li>&nbsp;The series is arithmetic. The sum does not exist.<\/li>\n \t<li>&nbsp;[latex]-\\dfrac{3}{11}[\/latex]<\/li>\n<\/ol>\n[\/hidden-answer]\n\n[ohm_question hide_question_numbers=1]20285[\/ohm_question]\n\n[ohm_question hide_question_numbers=1]20287[\/ohm_question]\n\n<\/div>\n<h2>Annuities<\/h2>\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.\n\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.\n\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,[\/latex] and [latex]n=72[\/latex] into the formula, and simplify to find the value of the annuity after 6 years.\n<p style=\"text-align: center;\">[latex]{S}_{72}=\\dfrac{50\\left(1-{1.005}^{72}\\right)}{1 - 1.005}\\approx 4\\text{,}320.44[\/latex]<\/p>\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.\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 \t<li>Determine [latex]{a}_{1}[\/latex], the value of the initial deposit.<\/li>\n \t<li>Determine [latex]n[\/latex], the number of deposits.<\/li>\n \t<li>Determine [latex]r[\/latex].\n<ol>\n \t<li>Divide the annual interest rate by the number of times per year that interest is compounded.<\/li>\n \t<li>Add 1 to this amount to find [latex]r[\/latex].<\/li>\n<\/ol>\n<\/li>\n \t<li>Substitute values for [latex]{a}_{1},r,[\/latex] and [latex]n[\/latex]\ninto the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, [latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n \t<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>\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?\n\n[reveal-answer q=\"808488\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"808488\"]\n\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.\n<p style=\"text-align: center;\">[latex]r=1+\\dfrac{0.09}{12}=1.0075[\/latex]<\/p>\nSubstitute [latex]{a}_{1}=100,r=1.0075,[\/latex] and [latex]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.\n<p style=\"text-align: center;\">[latex]{S}_{120}=\\dfrac{100\\left(1-{1.0075}^{120}\\right)}{1 - 1.0075}\\approx 19\\text{,}351.43[\/latex]<\/p>\nSo the account has $19,351.43 after the last deposit is made.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\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?\n\n[reveal-answer q=\"786342\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"786342\"]\n\n$92,408.18\n\n[\/hidden-answer]\n\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20277&amp;theme=oea&amp;iframe_resize_id=mom10[\/embed]\n\n<\/div>\n&nbsp;\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the first n terms of a geometric sequence.<\/li>\n<li class=\"li2\"><span class=\"s1\">Determine whether the sum of an in\ufb01nite geometric series exists.<\/span><\/li>\n<li>Give the sum of a convergent infinite geometric series.<\/li>\n<li class=\"li2\">Solve an annuity&nbsp;problem using a geometric series.<\/li>\n<\/ul>\n<\/div>\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}+{a}_{1}r+{a}_{1}{r}^{2}+...+{a}_{1}{r}^{n - 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}={a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+...+{a}_{1}{r}^{n}[\/latex]<\/p>\n<p>Next, we subtract this equation from the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&={a}_{1}+{a}_{1}r+{a}_{1}{r}^{2}+...+{a}_{1}{r}^{n - 1} \\\\ -r{S}_{n}&=-\\left({a}_{1}r+{a}_{1}{r}^{2}+{a}_{1}{r}^{3}+...+{a}_{1}{r}^{n}\\right) \\\\ \\hline \\left(1-r\\right){S}_{n}&={a}_{1}-{a}_{1}{r}^{n}\\end{align}[\/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], factor [latex]a_1[\/latex] on the right hand side and divide both sides by [latex]\\left(1-r\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>The formulas for finding the sum of the first [latex]n[\/latex] terms of arithmetic and geometric series are handy and should be memorized although it is important to understand how they were derived.<\/p>\n<\/div>\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}=\\dfrac{{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}=\\dfrac{{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]8 + -4 + 2 + \\dots[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^6 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\">Show 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=\\dfrac{-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{align}&{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &{S}_{11}=\\dfrac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336 \\\\ \\text{ } \\end{align}[\/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{align}\\\\ &{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\[1mm] &{S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378 \\end{align}[\/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]1\\text{,}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\">Show Solution<\/span><\/p>\n<div id=\"q922435\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\approx 2,000.00[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Use the formula to determine the sum&nbsp;[latex]\\sum\\limits _{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\">Show Solution<\/span><\/p>\n<div id=\"q15208\" class=\"hidden-answer\" style=\"display: none\">\n<p>9,840<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm19446\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=19446&#38;theme=oea&#38;iframe_resize_id=ohm19446&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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\">Show Solution<\/span><\/p>\n<div id=\"q636578\" class=\"hidden-answer\" style=\"display: none\">\n<p>The 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{align}{S}_{n}&=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{5}&=\\dfrac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03 \\end{align}[\/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\">Show Solution<\/span><\/p>\n<div id=\"q890801\" class=\"hidden-answer\" style=\"display: none\">\n<p>$275,513.31<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm5865\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5865&#38;theme=oea&#38;iframe_resize_id=ohm5865&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Formula for the Sum of an Infinite Geometric Series<\/h2>\n<p>Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first&nbsp;<em>n<\/em> terms. An&nbsp;<strong>infinite series<\/strong>&nbsp;is the sum of the terms of an infinite sequence. An example of an infinite series is [latex]2+4+6+8+\\dots[\/latex].<\/p>\n<p>This series can also be written in summation notation as [latex]\\sum\\limits _{k=1}^{\\infty} 2k[\/latex],&nbsp;where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series&nbsp;<strong>diverges<\/strong>.<\/p>\n<h3>Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h3>\n<p>If the terms of an&nbsp;<span class=\"no-emphasis\">infinite geometric series<\/span>&nbsp;approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:<\/p>\n<p style=\"text-align: center;\">[latex]1+0.2+0.04+0.008+0.0016+\\dots[\/latex]<\/p>\n<p>The common ratio is [latex]r=0.2[\/latex]. As&nbsp;<em>n<\/em> gets large, the values of of [latex]r^n[\/latex] get very small and approach 0.&nbsp;Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with [latex]-1<r<1[\/latex]&nbsp;approach 0; the sum of a geometric series is defined when&nbsp;[latex]-1<r<1[\/latex].\n\n\n<div class=\"textbox\">\n<h3>DETERMINING WHETHER THE SUM OF AN INFINITE GEOMETRIC SERIES IS DEFINED<\/h3>\n<p>The sum of an infinite series is defined if the series is geometric and&nbsp;[latex]-1<r<1[\/latex].\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: <strong>Given the first several terms of an infinite series, determine if the sum of the series exists.<\/strong><\/h3>\n<ol>\n<li>Find the ratio of the second term to the first term.<\/li>\n<li>Find the ratio of the third term to the second term.<\/li>\n<li>Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.<\/li>\n<li>If a common ratio,&nbsp;<em>r<\/em>, was found in step 3, check to see if&nbsp;[latex]-1<r<1[\/latex].&nbsp;If so, the sum is defined. If not, the sum is not defined.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Remember to try the examples below before looking at the answers!<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:&nbsp;Determining Whether the Sum of an Infinite Series is Defined<\/h3>\n<p>Determine whether the sum of each infinite series is defined.<\/p>\n<ol>\n<li>[latex]12+8+4+\\dots[\/latex]<\/li>\n<li>[latex]\\dfrac{3}{4}+\\dfrac{1}{2}+\\dfrac{1}{3}+\\dots[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty}{27}\\cdot\\left(\\dfrac{1}{3}\\right)^k[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty}{5k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q250515\">Show Solution<\/span><\/p>\n<div id=\"q250515\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The ratio of the second term to the first is [latex]\\frac{2}{3}[\/latex], which is not the same as the ratio of the third term to the second, [latex]\\frac{1}{2}[\/latex].&nbsp;The series is not geometric.<\/li>\n<li>The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of [latex]\\frac{2}{3}[\/latex]. The sum of the infinite series is defined.<\/li>\n<li>The given formula is exponential with a base of [latex]\\frac{1}{3}[\/latex]; the series is geometric with a common ratio of&nbsp;[latex]\\frac{1}{3}[\/latex]. The sum of the infinite series is defined.<\/li>\n<li>The given formula is not exponential. The series is arithmetic, not geometric and so cannot yield a finite sum.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Determine whether the sum of the infinite series is defined.<\/p>\n<ol>\n<li>[latex]\\dfrac{1}{3}+\\dfrac{1}{2}+\\dfrac{3}{4}+\\dfrac{9}{8}+\\cdots[\/latex]<\/li>\n<li>[latex]24+(-12)+6+(-3)+\\dots[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty} 15\\cdot(-0.3)^k[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q559520\">Show Solution<\/span><\/p>\n<div id=\"q559520\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The series is geometric, but [latex]r=\\dfrac{3}{2}>1[\/latex]. The sum is not defined.<\/li>\n<li>The series is geometric with [latex]r=-\\dfrac{1}{2}[\/latex]. The sum is defined.<\/li>\n<li>The series is geometric with [latex]r=-0.3[\/latex]. The sum is defined.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Sums of Infinite Series<\/h2>\n<p id=\"fs-id1165137679221\">When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first <em>n<\/em>&nbsp;terms of a geometric series.<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex]<\/p>\n<p>We will examine an infinite series with [latex]r=\\frac{1}{2}[\/latex]. What happens to [latex]r^n[\/latex] as&nbsp;<em>n<\/em> increases?<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &{\\left(\\frac{1}{2}\\right)}^{2} = \\frac{1}{4} \\\\&{\\left(\\frac{1}{2}\\right)}^{3} = \\frac{1}{8} \\\\&{\\left(\\frac{1}{2}\\right)}^{4} = \\frac{1}{16} \\end{align}[\/latex]<\/p>\n<p>The value of [latex]r^n[\/latex] decreases rapidly. What happens for greater values of&nbsp;<em>n<\/em>?<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &{\\left(\\frac{1}{2}\\right)}^{10} = \\frac{1}{1\\text{,}024} \\\\&{\\left(\\frac{1}{2}\\right)}^{20} = \\frac{1}{1\\text{,}048\\text{,}576} \\\\&{\\left(\\frac{1}{2}\\right)}^{30} = \\frac{1}{1\\text{,}073\\text{,}741\\text{,}824} \\end{align}[\/latex]<\/p>\n<p>As&nbsp;<em>n<\/em> gets large, [latex]r^n[\/latex] gets very small. We say that as&nbsp;<em>n<\/em> increases without bound,&nbsp;[latex]r^n[\/latex] approaches 0. As&nbsp;[latex]r^n[\/latex] approaches 0,&nbsp;[latex]1-r^n[\/latex] approaches 1. When this happens the numerator approaches [latex]a_1[\/latex]. This gives us the formula for the sum of an infinite geometric series.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>This is another very handy formula that should be memorized. It is important to understand, though, how it is derived from the formula for the&nbsp;sum of the first&nbsp;[latex]n[\/latex]&nbsp;terms of a geometric series for [latex]n[\/latex] increasing without bound.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: FORMULA FOR THE SUM OF AN INFINITE GEOMETRIC SERIES<\/h3>\n<p>The formula for the sum of an infinite geometric series with [latex]-1<r<1[\/latex] is:\n\n\n<p style=\"text-align: center;\">[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: <strong>Given an infinite geometric series, find its sum.<\/strong><\/h3>\n<ol>\n<li>Identify [latex]a_1[\/latex] and&nbsp;<em>r<\/em>.<\/li>\n<li>Confirm that [latex]-1<r<1[\/latex].<\/li>\n<li>Substitute values for&nbsp;[latex]a_1[\/latex] and&nbsp;<em>r<\/em> into the formula,&nbsp;[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex].<\/li>\n<li>Simplify to find&nbsp;<em>S<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:&nbsp;Finding the Sum of an Infinite Geometric Series<\/h3>\n<p>Find the sum, if it exists, for the following:<\/p>\n<ol>\n<li>[latex]10+9+8+7+\\dots[\/latex]<\/li>\n<li><span id=\"MJXp-Span-5192\" class=\"MJXp-mo\">[latex]248.6+99.44+39.776+\\dots[\/latex]<\/span><\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty}4\\text{,}374\\cdot\\left(-\\dfrac{1}{3}\\right)^{k-1}[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty}\\dfrac{1}{9}\\cdot\\left(\\dfrac{4}{3}\\right)^{k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q18513\">Show Solution<\/span><\/p>\n<div id=\"q18513\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>There is not a constant ratio; the series is not geometric.<\/li>\n<li>There is a constant ratio; the series is geometric. [latex]a_1=248.6[\/latex] and [latex]r=\\dfrac{99.44}{248.6}=0.4[\/latex], so the sum exists. Substitute&nbsp;[latex]a_1=248.6[\/latex] and [latex]r=0.4[\/latex] into the formula and simplify to find the sum.<br \/>\n[latex]\\begin{align} \\\\ &S=\\frac{a_1}{1-r} \\\\[1.5mm] &S=\\frac{248.6}{1-0.4}=\\frac{1243}{3} \\\\ \\text{ }\\end{align}[\/latex]<\/li>\n<li>The formula is exponential, so the series is geometric with [latex]r=-\\frac{1}{3}[\/latex]. Find [latex]a_1[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.<br \/>\n[latex]\\begin{align} \\\\ a_1=4\\text{,}374\\cdot\\left(-\\frac{1}{3}\\right)^{1-1}=4\\text{,}374 \\\\ \\text{ }\\end{align}[\/latex]<br \/>\nSubstitute [latex]4\\text{,}374[\/latex] and [latex]r=-\\frac{1}{3}[\/latex] into the formula, and simplify to find the sum.<br \/>\n[latex]\\begin{align}\\\\&S=\\frac{a_1}{1-r} \\\\[1.5mm] &S=\\frac{4\\text{,}374}{1-\\left(-\\frac{1}{3}\\right)}=3\\text{,}280.5 \\\\ \\text{ }\\end{align}[\/latex]<\/li>\n<li>The formula is exponential, so the series is geometric, but [latex]r>1[\/latex].&nbsp;The sum does not exist.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding an Equivalent Fraction for a Repeating Decimal<\/h3>\n<p>Find the equivalent fraction for the repeating decimal [latex]0.\\overline{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q624358\">Show Solution<\/span><\/p>\n<div id=\"q624358\" class=\"hidden-answer\" style=\"display: none\">\n<p>We notice the repeating decimal [latex]0.\\overline{3}=0.333\\dots[\/latex].&nbsp;so we can rewrite the repeating decimal as a sum of terms.<\/p>\n<p style=\"text-align: center;\">[latex]0.\\overline{3}=0.3+0.03+0.003+\\dots[\/latex]<\/p>\n<p>Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}0.\\overline{3}&=0.3+0.3\\cdot(0.1)+0.3\\cdot(0.01)+0.3\\cdot(0.001)+\\dots \\\\ &=0.3+0.3\\cdot(0.1)+0.3\\cdot(0.1)^2+0.3\\cdot(0.1)^3+\\dots\\end{align}[\/latex]<\/p>\n<p>Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have<\/p>\n<p style=\"text-align: center;\">[latex]S=\\dfrac{a_1}{1-r} =\\dfrac{0.3}{1-0.1} =\\dfrac{0.3}{0.9} =\\dfrac{1}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the sum if it exists.<\/p>\n<ol>\n<li>[latex]2+\\dfrac{2}{3}+\\dfrac{2}{9}+\\dots[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty}{0.76k+1}[\/latex]<\/li>\n<li>[latex]\\sum\\limits _{k=1}^{\\infty}\\left(-\\dfrac{3}{8}\\right)^k[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q221023\">Show Solution<\/span><\/p>\n<div id=\"q221023\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>&nbsp;3<\/li>\n<li>&nbsp;The series is arithmetic. The sum does not exist.<\/li>\n<li>&nbsp;[latex]-\\dfrac{3}{11}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm20285\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20285&theme=oea&iframe_resize_id=ohm20285\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm20287\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20287&theme=oea&iframe_resize_id=ohm20287\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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,[\/latex] and [latex]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}=\\dfrac{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},r,[\/latex] and [latex]n[\/latex]<br \/>\ninto the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, [latex]{S}_{n}=\\dfrac{{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\">Show 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+\\dfrac{0.09}{12}=1.0075[\/latex]<\/p>\n<p>Substitute [latex]{a}_{1}=100,r=1.0075,[\/latex] and [latex]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}=\\dfrac{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\">Show Solution<\/span><\/p>\n<div id=\"q786342\" class=\"hidden-answer\" style=\"display: none\">\n<p>$92,408.18<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm20277\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20277&#38;theme=oea&#38;iframe_resize_id=ohm20277&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-450\">\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":17533,"menu_order":19,"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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