{"id":2069,"date":"2015-11-12T18:30:42","date_gmt":"2015-11-12T18:30:42","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=2069"},"modified":"2017-04-03T19:34:20","modified_gmt":"2017-04-03T19:34:20","slug":"using-the-formula-for-arithmetic-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/using-the-formula-for-arithmetic-series\/","title":{"raw":"Using the Formula for Arithmetic Series","rendered":"Using the Formula for Arithmetic Series"},"content":{"raw":"

Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence<\/strong> is a sequence in which the difference between any two consecutive terms is the common difference<\/strong>, [latex]d[\/latex]. The sum of the terms of an arithmetic sequence is called an arithmetic series<\/strong>. We can write the sum of the first [latex]n[\/latex] terms of an arithmetic series as:\r\n<\/p>

[latex]{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}[\/latex].<\/div>\r\nWe can also reverse the order of the terms and write the sum as\r\n
[latex]{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}[\/latex].<\/div>\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
[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]<\/div>\r\nBecause there are [latex]n[\/latex] terms in the series, we can simplify this sum to\r\n
[latex]2{S}_{n}=n\\left({a}_{1}+{a}_{n}\\right)[\/latex].<\/div>\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
[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\r\n
\r\n

A General Note: Formula for the Sum of the First n<\/em> Terms of an Arithmetic Series<\/h3>\r\nAn arithmetic series<\/strong> is the sum of the terms of an arithmetic sequence. The formula for the sum of the first [latex]n[\/latex] terms of an arithmetic sequence is\r\n
[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\r\n<\/div>\r\n
\r\n

How To: Given terms of an arithmetic series, find the sum of the first [latex]n[\/latex] terms.<\/h3>\r\n
  1. Identify [latex]{a}_{1}[\/latex] and [latex]{a}_{n}[\/latex].<\/li>\r\n\t
  2. Determine [latex]n[\/latex].<\/li>\r\n\t
  3. 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
  4. Simplify to find [latex]{S}_{n}[\/latex].<\/li>\r\n<\/ol><\/div>\r\n
    \r\n

    Example 2: Finding the First n<\/em> Terms of an Arithmetic Series<\/h3>\r\nFind the sum of each arithmetic series.\r\n
    1. [latex]\\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[\/latex]<\/li>\r\n\t
    2. [latex]\\text{20 + 15 + 10 +}\\ldots{ + -50}[\/latex]<\/li>\r\n\t
    3. [latex]\\sum _{k=1}^{12}3k - 8[\/latex]<\/li>\r\n<\/ol><\/div>\r\n
      \r\n

      Solution<\/h3>\r\n
      1. 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].\r\n\r\nSubstitute 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]<\/div><\/li>\r\n\t
      2. 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]<\/div>\r\nSubstitute values for [latex]{a}_{1},{a}_{n}\\text{,}n[\/latex] into the formula and simplify.\r\n\r\n
        [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]<\/div><\/li>\r\n\t
      3. 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]<\/div>\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]<\/div>\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]<\/div><\/li>\r\n<\/ol><\/div>\r\nUse the formula to find the sum of each arithmetic series.\r\n
        \r\n

        Try It 2<\/h3>\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\nSolution<\/a>\r\n\r\n<\/div>\r\n
        \r\n

        Try It 3<\/h3>\r\n[latex]\\text{13 + 21 + 29 + }\\dots \\text{+ 69}[\/latex]\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
        \r\n

        Try It 4<\/h3>\r\n[latex]\\sum _{k=1}^{10}5 - 6k[\/latex]\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
        \r\n

        Example 3: 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<\/div>\r\n
        \r\n

        Solution<\/h3>\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
        [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]<\/div>\r\nWe can now use the formula for arithmetic series.\r\n
        [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]<\/div>\r\nShe will have walked a total of 11 miles.\r\n\r\n<\/div>\r\n
        \r\n

        Try It 5<\/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\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

        Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence<\/strong> is a sequence in which the difference between any two consecutive terms is the common difference<\/strong>, [latex]d[\/latex]. The sum of the terms of an arithmetic sequence is called an arithmetic series<\/strong>. We can write the sum of the first [latex]n[\/latex] terms of an arithmetic series as:\n<\/p>\n

        [latex]{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}[\/latex].<\/div>\n

        We can also reverse the order of the terms and write the sum as<\/p>\n

        [latex]{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}[\/latex].<\/div>\n

        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

        [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]<\/div>\n

        Because there are [latex]n[\/latex] terms in the series, we can simplify this sum to<\/p>\n

        [latex]2{S}_{n}=n\\left({a}_{1}+{a}_{n}\\right)[\/latex].<\/div>\n

        We divide by 2 to find the formula for the sum of the first [latex]n[\/latex] terms of an arithmetic series.<\/p>\n

        [latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\n
        \n

        A General Note: Formula for the Sum of the First n<\/em> Terms of an Arithmetic Series<\/h3>\n

        An arithmetic series<\/strong> is the sum of the terms of an arithmetic sequence. The formula for the sum of the first [latex]n[\/latex] terms of an arithmetic sequence is<\/p>\n

        [latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\n<\/div>\n
        \n

        How To: Given terms of an arithmetic series, find the sum of the first [latex]n[\/latex] terms.<\/h3>\n
          \n
        1. Identify [latex]{a}_{1}[\/latex] and [latex]{a}_{n}[\/latex].<\/li>\n
        2. Determine [latex]n[\/latex].<\/li>\n
        3. 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
        4. Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol>\n<\/div>\n
          \n

          Example 2: Finding the First n<\/em> Terms of an Arithmetic Series<\/h3>\n

          Find the sum of each arithmetic series.<\/p>\n

            \n
          1. [latex]\\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[\/latex]<\/li>\n
          2. [latex]\\text{20 + 15 + 10 +}\\ldots{ + -50}[\/latex]<\/li>\n
          3. [latex]\\sum _{k=1}^{12}3k - 8[\/latex]<\/li>\n<\/ol>\n<\/div>\n
            \n

            Solution<\/h3>\n
              \n
            1. 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].\n

              Substitute values for [latex]{a}_{1},{a}_{n}\\text{\\hspace{0.17em},}[\/latex] and [latex]n[\/latex] into the formula and simplify.<\/p>\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]<\/div>\n<\/li>\n
            2. 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].\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]<\/div>\n

              Substitute values for [latex]{a}_{1},{a}_{n}\\text{,}n[\/latex] into the formula and simplify.<\/p>\n

              \n
              [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]<\/div>\n<\/div>\n<\/li>\n
            3. To find [latex]{a}_{1}[\/latex], substitute [latex]k=1[\/latex] into the given explicit formula.\n
              [latex]\\begin{array}{l}{a}_{k}=3k - 8\\hfill \\\\ \\text{ }{a}_{1}=3\\left(1\\right)-8=-5\\hfill \\end{array}[\/latex]<\/div>\n

              We are given that [latex]n=12[\/latex]. To find [latex]{a}_{12}[\/latex], substitute [latex]k=12[\/latex] into the given explicit formula.<\/p>\n

              [latex]\\begin{array}{l}\\text{ }{a}_{k}=3k - 8\\hfill \\\\ {a}_{12}=3\\left(12\\right)-8=28\\hfill \\end{array}[\/latex]<\/div>\n

              Substitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.<\/p>\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]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

              Use the formula to find the sum of each arithmetic series.<\/p>\n

              \n

              Try It 2<\/h3>\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]<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n

              \n

              Try It 3<\/h3>\n

              [latex]\\text{13 + 21 + 29 + }\\dots \\text{+ 69}[\/latex]<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n

              \n

              Try It 4<\/h3>\n

              [latex]\\sum _{k=1}^{10}5 - 6k[\/latex]<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n

              \n

              Example 3: Solving Application Problems with Arithmetic Series<\/h3>\n

              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>\n

              \n

              Solution<\/h3>\n

              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

              [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]<\/div>\n

              We can now use the formula for arithmetic series.<\/p>\n

              [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]<\/div>\n

              She will have walked a total of 11 miles.<\/p>\n<\/div>\n

              \n

              Try It 5<\/h3>\n

              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

              Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

              \n\t\t\t

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