{"id":2053,"date":"2015-11-12T18:30:42","date_gmt":"2015-11-12T18:30:42","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=2053"},"modified":"2015-11-12T18:30:42","modified_gmt":"2015-11-12T18:30:42","slug":"finding-common-ratios","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/finding-common-ratios\/","title":{"raw":"Finding Common Ratios","rendered":"Finding Common Ratios"},"content":{"raw":"

The yearly salary values described form a geometric sequence<\/strong> because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio<\/strong>. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.\n\n\"A<\/p>

\n

A General Note: Definition of a Geometric Sequence<\/h3>\nA geometric sequence<\/strong> is one in which any term divided by the previous term is a constant. This constant is called the common ratio<\/strong> of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If [latex]{a}_{1}[\/latex] is the initial term of a geometric sequence and [latex]r[\/latex] is the common ratio, the sequence will be\n
[latex]\\left\\{{a}_{1}, {a}_{1}r,{a}_{1}{r}^{2},{a}_{1}{r}^{3},...\\right\\}[\/latex].<\/div>\n<\/div>\n
\n

How To: Given a set of numbers, determine if they represent a geometric sequence.<\/h3>\n
  1. Divide each term by the previous term.<\/li>\n\t
  2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.<\/li>\n<\/ol><\/div>\n
    \n

    Example 1: Finding Common Ratios<\/h3>\nIs the sequence geometric? If so, find the common ratio.\n
    1. [latex]1\\text{,}2\\text{,}4\\text{,}8\\text{,}16\\text{,}..[\/latex].<\/li>\n\t
    2. [latex]48\\text{,}12\\text{,}4\\text{, }2\\text{,}..[\/latex].<\/li>\n<\/ol><\/div>\n
      \n

      Solution<\/h3>\nDivide each term by the previous term to determine whether a common ratio exists.\n
      1. [latex]\\begin{array}{llllllllll}\\frac{2}{1}=2\\hfill & \\hfill & \\hfill & \\frac{4}{2}=2\\hfill & \\hfill & \\hfill & \\frac{8}{4}=2\\hfill & \\hfill & \\hfill & \\frac{16}{8}=2\\hfill \\end{array}[\/latex]The sequence is geometric because there is a common ratio. The common ratio is 2.<\/li>\n\t
      2. [latex]\\begin{array}{lllllll}\\frac{12}{48}=\\frac{1}{4}\\hfill & \\hfill & \\hfill & \\frac{4}{12}=\\frac{1}{3}\\hfill & \\hfill & \\hfill & \\frac{2}{4}=\\frac{1}{2}\\hfill \\end{array}[\/latex]The sequence is not geometric because there is not a common ratio.<\/li>\n<\/ol><\/div>\n
        \n

        Analysis of the Solution<\/h3>\nThe graph of each sequence is shown in Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"GraphFigure 1<\/b>[\/caption]\n\n<\/div>\n
        \n

        Q & A<\/h3>\n

        If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?<\/h3>\nNo. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.<\/em>\n\n<\/div>\n
        \n

        Try It 1<\/h3>\nIs the sequence geometric? If so, find the common ratio.\n
        [latex]5,10,15,20,..[\/latex].<\/div>\n\nSolution<\/a>\n\n<\/div>\n
        \n

        Try It 2<\/h3>\nIs the sequence geometric? If so, find the common ratio.\n
        [latex]100,20,4,\\frac{4}{5},..[\/latex].<\/div>\n\nSolution<\/a>\n\n<\/div>","rendered":"

        The yearly salary values described form a geometric sequence<\/strong> because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio<\/strong>. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term.<\/p>\n

        \"A<\/p>\n

        \n

        A General Note: Definition of a Geometric Sequence<\/h3>\n

        A geometric sequence<\/strong> is one in which any term divided by the previous term is a constant. This constant is called the common ratio<\/strong> of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If [latex]{a}_{1}[\/latex] is the initial term of a geometric sequence and [latex]r[\/latex] is the common ratio, the sequence will be<\/p>\n

        [latex]\\left\\{{a}_{1}, {a}_{1}r,{a}_{1}{r}^{2},{a}_{1}{r}^{3},...\\right\\}[\/latex].<\/div>\n<\/div>\n
        \n

        How To: Given a set of numbers, determine if they represent a geometric sequence.<\/h3>\n
          \n
        1. Divide each term by the previous term.<\/li>\n
        2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric.<\/li>\n<\/ol>\n<\/div>\n
          \n

          Example 1: Finding Common Ratios<\/h3>\n

          Is the sequence geometric? If so, find the common ratio.<\/p>\n

            \n
          1. [latex]1\\text{,}2\\text{,}4\\text{,}8\\text{,}16\\text{,}..[\/latex].<\/li>\n
          2. [latex]48\\text{,}12\\text{,}4\\text{, }2\\text{,}..[\/latex].<\/li>\n<\/ol>\n<\/div>\n
            \n

            Solution<\/h3>\n

            Divide each term by the previous term to determine whether a common ratio exists.<\/p>\n

              \n
            1. [latex]\\begin{array}{llllllllll}\\frac{2}{1}=2\\hfill & \\hfill & \\hfill & \\frac{4}{2}=2\\hfill & \\hfill & \\hfill & \\frac{8}{4}=2\\hfill & \\hfill & \\hfill & \\frac{16}{8}=2\\hfill \\end{array}[\/latex]The sequence is geometric because there is a common ratio. The common ratio is 2.<\/li>\n
            2. [latex]\\begin{array}{lllllll}\\frac{12}{48}=\\frac{1}{4}\\hfill & \\hfill & \\hfill & \\frac{4}{12}=\\frac{1}{3}\\hfill & \\hfill & \\hfill & \\frac{2}{4}=\\frac{1}{2}\\hfill \\end{array}[\/latex]The sequence is not geometric because there is not a common ratio.<\/li>\n<\/ol>\n<\/div>\n
              \n

              Analysis of the Solution<\/h3>\n

              The graph of each sequence is shown in Figure 1. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not.<\/p>\n

              \"Graph<\/p>\n

              Figure 1<\/b><\/p>\n<\/div>\n<\/div>\n

              \n

              Q & A<\/h3>\n

              If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio?<\/h3>\n

              No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio.<\/em><\/p>\n<\/div>\n

              \n

              Try It 1<\/h3>\n

              Is the sequence geometric? If so, find the common ratio.<\/p>\n

              [latex]5,10,15,20,..[\/latex].<\/div>\n

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

              \n

              Try It 2<\/h3>\n

              Is the sequence geometric? If so, find the common ratio.<\/p>\n

              [latex]100,20,4,\\frac{4}{5},..[\/latex].<\/div>\n

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

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