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24.1: Finite Geometric Series

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    We now study another sequence, the geometric sequence, which will be analogous to our study of the arithmetic sequence in section 23.2. We have already encountered examples of geometric sequences in Example 23.1.1 (b). A geometric sequence is a sequence for which we multiply a constant number to get from one term to the next, for example:

    \[5,\underset{\times 4}{\hookrightarrow } 20,\underset{\times 4}{\hookrightarrow } 80,\underset{\times 4}{\hookrightarrow } 320,\underset{\times 4}{\hookrightarrow } 1280, \dots \nonumber \]

    Definition: Geometric Sequence

    A sequence \(\{a_n\}\) is called a geometric sequence, if any two consecutive terms have a common ratio \(r\). The geometric sequence is determined by \(r\) and the first value \(a_1\). This can be written recursively as:

    \[a_n=a_{n-1}\cdot r \quad \quad \text{for }n\geq 2 \nonumber \]

    Alternatively, we have the general formula for the \(n\)th term of the geometric sequence:

    \[\label{EQU:geometric-sequence-general-term} \boxed{a_n=a_1\cdot r^{n-1}}\]

    Example \(\PageIndex{1}\)

    Determine if the sequence is a geometric, or arithmetic sequence, or neither or both. If it is a geometric or arithmetic sequence, then find the general formula for \(a_n\) in the form \(\ref{EQU:geometric-sequence-general-term}\) or [EQU:arithmetic-sequence-general-term].

    1. \(3, 6, 12, 24, 48, \dots\)
    2. \(100, 50, 25, 12.5, \dots\)
    3. \(700, -70, 7, -0.7, 0.07, \dots\)
    4. \(2, 4, 16, 256, \dots\)
    5. \(3, 10, 17, 24, \dots\)
    6. \(-3, -3, -3, -3, -3, \dots\)
    7. \(a_n=\left(\dfrac{3}{7}\right)^n\)
    8. \(a_n=n^2\)

    Solution

    1. Calculating the quotient of two consecutive terms always gives the same number \(6\div 3=2\), \(12\div 6=2\), \(24\div 12=2\), etc. Therefore the common ratio is \(r=2\), which shows that this is a geometric sequence. Furthermore, the first term is \(a_1=3\), so that the general formula for the \(n\)th term is \(a_n=3\cdot 2^{n-1}\).
    2. We see that the common ratio between two terms is \(r=\dfrac 1 2\), so that this is a geometric sequence. Since the first term is \(a_1=100\), we have the general term \(a_n=100\cdot \left(\dfrac 1 2\right)^{n-1}\).
    3. Two consecutive terms have a ratio of \(r=-\dfrac 1 {10}\), and the first term is \(a_1=700\). The general term of this geometric sequence is \(a_n=700\cdot \left(-\dfrac 1 {10}\right)^{n-1}\).
    4. The quotient of the first two terms is \(4\div 2=2\), whereas the quotient of the next two terms is \(16\div 4=4\). Since these quotients are not equal, this is not a geometric sequence. Furthermore, the difference between the first two terms is \(4-2=2\), and the next two terms have a difference \(16-4=12\). Therefore, this is also not an arithmetic sequence.
    5. The quotient of the first couple of terms is not equal \(\dfrac{10}{3}\neq \dfrac{17}{10}\), so that this is not a geometric sequence. The difference of any two terms is \(7=10-3=17-10=24-17\), so that this is part of an arithmetic sequence with common difference \(d=7\). The general formula is \(a_n=a_1+d\cdot (n-1)=3+7\cdot (n-1)\).
    6. The common ratio is \(r=(-3)\div (-3)=1\), so that this is a geometric sequence with \(a_n=(-3)\cdot 1^{n-1}\). On the other hand, the common difference is \((-3)-(-3)=0\), so that this is also an arithmetic sequence with \(a_n=(-3)+0\cdot (n-1)\). Of course, both formulas reduce to the simpler expression \(a_n=-3\).
    7. Writing the first couple of terms in the sequence \(\left\{\left(\dfrac 3 7 \right)^n\right\}\), we obtain:

    \[\left(\dfrac 3 7 \right)^1, \left(\dfrac 3 7 \right)^2, \left(\dfrac 3 7 \right)^3, \left(\dfrac 3 7 \right)^4, \left(\dfrac 3 7 \right)^5, \dots \nonumber \]

    Thus, we get from one term to the next by multiplying \(r=\dfrac 3 7\), so that this is a geometric sequence. The first term is \(a_1=\dfrac 3 7\), so that \(a_n=\dfrac 3 7 \cdot \left(\dfrac 3 7 \right)^{n-1}\). This is clearly the given sequence, since we may simplify this as

    \[a_n=\dfrac 3 7 \cdot \left(\dfrac 3 7 \right)^{n-1} =\left(\dfrac 3 7 \right)^{1+n-1}=\left(\dfrac 3 7 \right)^{n} \nonumber \]

    1. We write the first terms in the sequence \(\{n^2\}_{n\geq 1}\):

    \[1, 4, 9, 16, 25, 36, 49, \dots \nonumber \]

    Calculating the quotients of consecutive terms, we get \(4\div 1 =4\) and \(9\div 4=2.25\), so that this is not a geometric sequence. Also the difference of consecutive terms is \(4-1=3\) and \(9-4=5\), so that this is also not an arithmetic sequence.

    Example \(\PageIndex{2}\)

    Find the general formula of a geometric sequence with the given property.

    1. \(r=4\), and \(a_5=6400\)
    2. \(a_1=\dfrac{2}{5}\), and \(a_{4}=-\dfrac{27}{20}\)
    3. \(a_{5}=216\), \(a_{7}=24\), and \(r\) is positive

    Solution

    1. Since \(\{a_n\}\) is a geometric sequence, it is \(a_n=a_1\cdot r^{n-1}\). We know that \(r=4\), so we still need to find \(a_1\). Using \(a_5=64000\), we obtain:

    \[6400=a_5=a_1\cdot 4^{5-1}=a_1\cdot 4^4=256\cdot a_1 \,\, \stackrel{(\div 256)}{\implies} \,\, a_1=\dfrac{6400}{256}=25 \nonumber \]

    The sequence is therefore given by the formula, \(a_n=25\cdot 4^{n-1}\).

    1. The geometric sequence \(a_n=a_1\cdot r^{n-1}\) has \(a_1=\dfrac{2}{5}\). We calculate \(r\) using the second condition.

    \[\begin{aligned} -\dfrac{27}{20}=a_4=a_1\cdot r^{4-1}=\dfrac{2}{5}\cdot r^3 & \stackrel{(\times \frac{5}{2})}{\implies} & r^3= -\dfrac{27}{20}\cdot \dfrac 5 2= -\dfrac{27}{4}\cdot \dfrac 1 2=\dfrac{-27}8 \\ & \stackrel{(\text{take }\sqrt[3]{\,\,})}{\implies} & r=\sqrt[3]{\dfrac{-27}8}=\dfrac{\sqrt[3]{-27}}{\sqrt[3]{8}}=\dfrac{-3}{2}\end{aligned}\nonumber \]

    Therefore, \(a_n=\dfrac 2 5 \cdot \left(\dfrac {-3} 2\right)^{n-1}\).

    1. The question does neither provide \(a_1\) nor \(r\) in the formula \(a_n=a_1\cdot r^{n-1}\). However, we obtain two equations in the two variables \(a_1\) and \(r\):

    \[\left\{\begin{array} { c }
    { 2 1 6 = a _ { 5 } = a _ { 1 } \cdot r ^ { 5 - 1 } } \\
    { 2 4 = a _ { 7 } = a _ { 1 } \cdot r ^ { 7 - 1 } }
    \end{array} \quad \Longrightarrow \left\{\begin{array}{c}
    216=a_{1} \cdot r^{4} \\
    24=a_{1} \cdot r^{6}
    \end{array}\right.\right. \nonumber \]

    In order to solve this, we need to eliminate one of the variables. Looking at the equations on the right, we see dividing the top equation by the bottom equation cancels \(a_1\).

    \[\dfrac{216}{24}=\dfrac{a_1\cdot r^4}{a_1\cdot r^6} \,\, \implies \,\, \dfrac{9}{1}=\dfrac{1}{r^2} \,\,\stackrel{(\text{take reciprocal})}\implies \,\, \dfrac{1}{9}=\dfrac{r^2}{1} \,\, \implies \,\, r^2= \dfrac 1 9 \nonumber \]

    To obtain \(r\) we have to solve this quadratic equation. In general, there are in fact two solutions:

    \[r=\pm\sqrt{\dfrac 1 9}=\pm\dfrac 1 3 \nonumber \]

    Since the problem states that \(r\) is positive, we see that we need to take the positive solution \(r=\dfrac 1 3\). Plugging \(r=\dfrac 1 3\) back into either of the two equations, we may solve for \(a_1\). For example, using the first equation \(a_5=216\), we obtain:

    \[\begin{aligned} && 216=a_5=a_1\cdot \bigg(\dfrac 1 3\bigg)^{5-1}=a_1\cdot \bigg(\dfrac 1 3\bigg)^{4}=a_1\cdot \dfrac 1 {3^4}=a_1\cdot \dfrac 1 {81} \\ && \quad \stackrel{(\times 81)}\implies \quad a_1=81\cdot 216 = 17496\end{aligned} \nonumber \]

    So, we finally arrive at the general formula for the \(n\)th term of the geometric sequence, \(a_n=17496\cdot \left(\dfrac 1 3\right)^{n-1}\).

    We can find the sum of the first \(k\) terms of a geometric sequence using another trick, which is very different from the one we used for the arithmetic sequence.

    Example \(\PageIndex{3}\)

    Consider the geometric sequence \(a_n=8\cdot 5^{n-1}\), that is the sequence:

    \[8, 40, 200, 1000, 5000, 25000, 125000, \dots \nonumber \]

    We want to add the first \(6\) terms of this sequence. \[8+ 40+ 200+ 1000+ 5000+ 25000=31248 \nonumber \]

    Solution

    In general, it may be much more difficult to simply add the terms as we did above, and we need to use a better general method. For this, we multiply \((1-5)\) to the sum \((8+ 40+ 200+ 1000+ 5000+ 25000)\) and simplify this using the distributive law:

    \[\begin{aligned} (1-5)\cdot (8+ 40+ 200+ 1000+ 5000+ 25000) &= 8-40+ 40-200+ 200-1000+ 1000-5000 + 5000-25000+ 25000-125000\\ &= 8-125000 \end{aligned} \nonumber \]

    In the second and third lines above, we have what is called a telescopic sum, which can be canceled except for the very first and last terms. Dividing by \((1-5)\), we obtain:

    \[8+ 40+ 200+ 1000+ 5000+ 25000=\dfrac{8-125000}{1-5}=\dfrac{-124992}{-4}=31248 \nonumber \]

    The previous example generalizes to the more general setting starting with an arbitrary geometric sequence.

    Observation: Geometric series

    Let \(\{a_n\}\) be a geometric sequence, whose \(n\)th term is given by the formula \(a_n=a_1\cdot r^{n-1}\). We furthermore assume that \(r\neq 1\). Then, the sum \(a_1+a_2+\dots+a_{k-1}+a_k\) is given by:

    \[\label{EQU:geometric-series} \boxed{\sum_{i=1}^k a_i =a_1\cdot \dfrac{1-r^k}{1-r}}\]

    Proof

    We multiply \((1-r)\) to the sum \((a_1+a_2+\dots+a_{k-1}+a_k)\):

    \[\begin{aligned} (1-r)\cdot (a_1+a_2+\dots+a_k) &= (1-r)\cdot (a_1 \cdot r^0+a_1\cdot r^1+\dots+a_1\cdot r^{k-1}) \\ &= a_1 \cdot r^0-a_1\cdot r^1+a_1\cdot r^1-a_1\cdot r^2+\dots+a_1 \cdot r^{k-1}-a_1\cdot r^k \\ &= a_1\cdot r^0-a_1\cdot r^k \\&= a_1\cdot (1-r^k)\end{aligned} \nonumber \]

    Dividing by \((1-r)\), we obtain

    \[a_1+a_2+\dots+a_k=\dfrac{a_1\cdot (1-r^k)}{(1-r)}=a_1\cdot \dfrac{1-r^k}{1-r} \nonumber \]

    This is the formula we wanted to prove.

    Example \(\PageIndex{4}\)

    Find the value of the geometric series.

    1. Find the sum \(\sum\limits_{n=1}^6 a_n\) for the geometric sequence \(a_n=10\cdot 3^{n-1}\).
    2. Determine the value of the geometric series: \(\sum\limits_{k=1}^{5} \left(-\dfrac{1}{2}\right)^k\)
    3. Find the sum of the first \(12\) terms of the geometric sequence \[-3, -6, -12, -24, \dots \nonumber\]

    Solution

    1. We need to find the sum \(a_1+a_2+a_3+a_4+a_5+a_6\), and we will do so using the formula provided in equation \(\ref{EQU:geometric-series}\). Since \(a_n=10\cdot 3^{n-1}\), we have \(a_1=10\) and \(r=3\), so that

    \[\sum_{n=1}^6 a_n = 10\cdot \dfrac{1-3^6}{1-3} = 10\cdot \dfrac{1-729}{1-3}=10\cdot \dfrac{-728}{-2}=10\cdot 364=3640 \nonumber \]

    1. Again, we use the formula for the geometric series \(\sum_{k=1}^n a_k=a_1\cdot \dfrac{1-r^n}{1-r}\), since \(a_k=\left(-\dfrac 1 2\right)^k\) is a geometric series. We may calculate the first term \(a_1=-\dfrac 1 2\), and the common ratio is also \(r=-\dfrac 1 2\). With this, we obtain:

    \[\begin{aligned} \sum_{k=1}^{5} \left(-\dfrac 1 2\right)^k & = \left(-\dfrac 1 2\right) \cdot \dfrac{1-(-\frac 1 2)^5}{1-(-\frac 1 2)}\\& =\left(-\dfrac 1 2\right) \cdot \dfrac{1-\left((-1)^5 \frac {1^5} {2^5}\right)}{1-(-\frac 1 2)}\\&= \left(-\dfrac 1 2\right) \cdot \dfrac{1-(- \frac {1} {32})}{1-(-\frac 1 2)}\\& = \left(-\dfrac 1 2\right) \cdot \dfrac{1+ \frac {1} {32}}{1+\frac 1 2}\\& = \left(-\dfrac 1 2\right) \cdot \dfrac{ \frac {32+1} {32}}{\frac {2+1} 2} \\ &= \left(-\dfrac 1 2\right) \cdot \dfrac{ \frac {33} {32}}{\frac {3} 2} \\&= \left(-\dfrac 1 2\right) \cdot \dfrac {33} {32}\cdot \dfrac 2 3 \\&= -\dfrac 1 2 \cdot \dfrac {11} {16}\\&=-\dfrac {11}{32}\end{aligned} \nonumber \]

    1. Our first task is to find the formula for the provided geometric series \(-3, -6, -12, -24, \dots\). The first term is \(a_1=-3\) and the common ratio is \(r=2\), so that \(a_n=(-3)\cdot 2^{n-1}\). The sum of the first \(12\) terms of this sequence is again given by equation \(\ref{EQU:geometric-series}\):

    \[\begin{aligned} \sum_{i=1}^{12} (-3)\cdot 2^{i-1}&=(-3)\cdot \dfrac{1-2^{12}}{1-2}\\&=(-3)\cdot \dfrac{1-4096}{1-2}\\&=(-3)\cdot \dfrac{-4095}{-1} \\ &= (-3)\cdot 4095\\& = -12285 \end{aligned} \nonumber \]


    This page titled 24.1: Finite Geometric Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.