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Sums of Finite Geometric Series

Series with defined ending value has sum: Sn = (a1(1-r^n))/(1-r)

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Finding the Sum of a Finite Geometric Series

You are saving for summer camp. You deposit $100 on the first of each month into your savings account. The account grows at a rate of 0.5% per month. How much money is in your account on the first day on the \begin{align*}9^{th}\end{align*} month?

Sum of Finite Geometric Series

We have discussed how to use the calculator to find the sum of any series provided we know the \begin{align*}n^{th}\end{align*} term rule. For a geometric series, however, there is a specific rule that can be used to find the sum algebraically. Let’s look at a finite geometric sequence and derive this rule.

Given \begin{align*}a_n=a_1r^{n-1}\end{align*}.

The sum of the first \begin{align*}n\end{align*} terms of a geometric sequence is: \begin{align*}S_n=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1}\end{align*}.

Now, factor out \begin{align*}a_1\end{align*} to get \begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1})\end{align*}. If we isolate what is in the parenthesis and multiply this sum by \begin{align*}(1-r)\end{align*} as shown below we can simplify the sum:

\begin{align*}&(1-r)S_n=(1-r)(1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= (1-r)^n\end{align*}

By multiplying the sum by \begin{align*}1-r\end{align*} we were able to cancel out all of the middle terms. However, we have changed the sum by a factor of \begin{align*}1-r\end{align*}, so what we really need to do is multiply our sum by \begin{align*}\frac{1-r}{1-r}\end{align*}, or 1.

\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \frac{1-r}{1-r}=\frac{a_1(1-r^n)}{1-r}\end{align*}, which is the sum of a finite geometric series.

So, \begin{align*}S_n=\frac{a_1(1-r^n)}{1-r}\end{align*}.

Let's find the sum of the first ten terms of the geometric sequence \begin{align*}a_n=\frac{1}{32}(-2)^{n-1}\end{align*}. This could also be written as, “Let's find \begin{align*}\sum\limits_{n=1}^{10} \frac{1}{32}(-2)^{n-1}\end{align*}.”

Using the formula, \begin{align*}a_1=\frac{1}{32}\end{align*}, \begin{align*}r=-2\end{align*}, and \begin{align*}n=10\end{align*}.

\begin{align*}S_{10}=\frac{\frac{1}{32}(1-(-2)^{10})}{1-(-2)} = \frac{\frac{1}{32}(1-1024)}{3} = -\frac{341}{32}\end{align*}

We can also use the calculator as shown below.

\begin{align*}sum(seq(1/32(-2)^{x-1}, x, 1, 10))=-\frac{341}{32}\end{align*}

Now, let's find the first term and the \begin{align*}n^{th}\end{align*} term rule for a geometric series in which the sum of the first 5 terms is 242 and the common ratio is 3.

Plug in what we know to the formula for the sum and solve for the first term:

\begin{align*}242 &= \frac{a_1(1-3^5)}{1-3} \\ 242 &= \frac{a_1(-242)}{-2} \\ 242 &= 121a_1 \\ a_1 &= 2\end{align*}

The first term is 2 and \begin{align*}a_n=2(3)^{n-1}\end{align*}.

Finally, let's solve the following problem.

Charlie deposits $1000 on the first of each year into his investment account. The account grows at a rate of 8% per year. How much money is in the account on the first day on the \begin{align*}11^{th}\end{align*} year.

First, consider what is happening here on the first day of each year. On the first day of the first year, $1000 is deposited. On the first day of the second year $1000 is deposited and the previously deposited $1000 earns 8% interest or grows by a factor of 1.08 (108%). On the first day of the third year another $1000 is deposited, the previous year’s deposit earns 8% interest and the original deposit earns 8% interest for two years (we multiply by \begin{align*}1.08^2\end{align*}):

\begin{align*}&\text{Sum Year} \ 1: 1000\\ &\text{Sum Year} \ 2:1000 + 1000(1.08)\\ &\text{Sum Year} \ 3: 1000 + 1000(1.08) + 1000(1.08)^2\\ &\text{Sum Year} \ 4: 1000 + 1000(1.08) + 1000(1.08)^2 + 1000(1.08)^3\\ & \qquad \qquad \qquad \qquad \qquad \vdots\\ &\text{Sum Year} \ 11: 1000 + 1000(1.08) + 1000(1.08)^2 + 1000(1.08)^3 + \ldots + 1000(1.08)^9 + 1000(1.08)^{10}\end{align*}

\begin{align*}^*\end{align*} There are 11 terms in this series because on the first day of the \begin{align*}11^{th}\end{align*} year we make our final deposit and the original deposit earns interest for 10 years.

This series is geometric. The first term is 1000, the common ratio is 1.08 and \begin{align*}n=11\end{align*}. Now we can calculate the sum using the formula and determine the value of the investment account at the start of the \begin{align*}11^{th}\end{align*} year.

\begin{align*}s_{11}=\frac{1000 \left(1-1.08^{11}\right)}{1-1.08}=16645.48746 \approx \$16,645.49\end{align*}

Examples

Example 1

Earlier, you were asked to find how much money is in your account on the first day of the \begin{align*}9^{th}\end{align*} month. 

There are 9 terms in this series because on the first day of the \begin{align*}9^{th}\end{align*} month you make your final deposit and the original deposit earns interest for 8 months.

This series is geometric. The first term is 100, the common ratio is 1.005 and \begin{align*}n=9\end{align*}. Now we can calculate the sum using the formula and determine the value of the investment account at the start of the \begin{align*}9^{th}\end{align*} month.

\begin{align*}s_{9}=\frac{100 \left(1-1.005^{9}\right)}{1-1.005}= 918.22\end{align*}

Therefore there is $918.22 in the account at the beginning of the ninth month.

Example 2

Evaluate \begin{align*}\sum\limits_{n=3}^8 2(-3)^{n-1}\end{align*}.

Since we are asked to find the sum of the \begin{align*}3^{rd}\end{align*} through \begin{align*}8^{th}\end{align*} terms, we will consider \begin{align*}a_3\end{align*} as the first term. The third term is \begin{align*}a_3=2(-3)^2=2(9)=18\end{align*}. Since we are starting with term three, we will be summing 6 terms, \begin{align*}a_3+a_4+a_5+a_6+a_7+a_8\end{align*}, in total. We can use the rule for the sum of a geometric series now with \begin{align*}a_1=18\end{align*}, \begin{align*}r=-3\end{align*} and \begin{align*}n=6\end{align*} to find the sum:

\begin{align*}\sum_{n=3}^8 2(-3)^{n-1}=\frac{18(1-(-3)^6)}{1-(-3)}=-3276\end{align*}

Example 3

If the sum of the first seven terms in a geometric series is \begin{align*}\frac{215}{8}\end{align*} and \begin{align*}r=-\frac{1}{2}\end{align*}, find the first term and the \begin{align*}n^{th}\end{align*} term rule.

We can substitute what we know into the formula for the sum of a geometric series and solve for \begin{align*}a_1\end{align*}.

\begin{align*}\frac{215}{8} &= \frac{a_1 \left(1-\left(-\frac{1}{2}\right)^7\right)}{1- \left(-\frac{1}{2}\right)} \\ \frac{215}{8} &= a_1 \left(\frac{43}{64}\right) \\ a_1 &= \left(\frac{64}{43}\right)\left(\frac{215}{8}\right)=40\end{align*}

The \begin{align*}n^{th}\end{align*} term rule is \begin{align*}a_n=40 \left(-\frac{1}{2}\right)^{n-1}\end{align*}

Example 4

Sam deposits $50 on the first of each month into an account which earns 0.5% interest each month. To the nearest dollar, how much is in the account right after Sam makes his last deposit on the first day of the fifth year (the \begin{align*}49^{th}\end{align*} month).

The deposits that Sam make and the interest earned on each deposit generate a geometric series,

\begin{align*}& S_{49}=50+50(1.005)^1+50(1.005)^2+50(1.005)^3+ \ldots +50(1.005)^{47}+50(1.005)^{48}, \\ & \qquad \ \ \uparrow \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \uparrow \\ & \quad \text{last deposit} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{first deposit}\end{align*}

Note that the first deposit earns interest for 48 months and the final deposit does not earn any interest. Now we can find the sum using \begin{align*}a_1=50\end{align*}, \begin{align*}r=1.005\end{align*} and \begin{align*}n=49\end{align*}.

\begin{align*}S_{49}=\frac{50(1-(1.005)^{49})}{(1-1.005)} \approx \$2768\end{align*}

Review

Use the formula for the sum of a geometric series to find the sum of the first five terms in each series.

  1. \begin{align*}a_n=36 \left(\frac{2}{3}\right)^{n-1}\end{align*}
  2. \begin{align*}a_n=9(-2)^{n-1}\end{align*}
  3. \begin{align*}a_n=5(-1)^{n-1}\end{align*}
  4. \begin{align*}a_n=\frac{8}{25} \left(\frac{5}{2}\right)^{n-1}\end{align*}
  5. \begin{align*}a_n=\frac{2}{3} \left(-\frac{3}{4}\right)^{n-1}\end{align*}

Find the indicated sums using the formula and then check your answers with the calculator.

  1. \begin{align*}\sum\limits_{n=1}^4(-1) \left(\frac{1}{2}\right)^{n-1}\end{align*}
  2. \begin{align*}\sum\limits_{n=2}^8(128) \left(\frac{1}{4}\right)^{n-1}\end{align*}
  3. \begin{align*}\sum\limits_{n=2}^7 \frac{125}{64}\left(\frac{4}{5}\right)^{n-1}\end{align*}
  4. \begin{align*}\sum\limits_{n=5}^{11} \frac{1}{32}(-2)^{n-1}\end{align*}

Given the sum and the common ratio, find the \begin{align*}n^{th}\end{align*} term rule for the series.

  1. \begin{align*}\sum\limits_{n=1}^6 a_n=-63\end{align*} and \begin{align*}r=-2\end{align*}
  2. \begin{align*}\sum\limits_{n=1}^4 a_n=671\end{align*} and \begin{align*}r=\frac{5}{6}\end{align*}
  3. \begin{align*}\sum\limits_{n=1}^5 a_n=122\end{align*} and \begin{align*}r=-3\end{align*}
  4. \begin{align*}\sum\limits_{n=2}^7 a_n=-\frac{63}{2}\end{align*} and \begin{align*}r=-\frac{1}{2}\end{align*}

Solve the following word problems using the formula for the sum of a geometric series.

  1. Sapna’s grandparents deposit $1200 into a college savings account on her \begin{align*}5^{th}\end{align*} birthday. They continue to make this birthday deposit each year until making the final deposit on her \begin{align*}18^{th}\end{align*} birthday. If the account earns 5% interest annually, how much is there after the final deposit?
  2. Jeremy wants to have save $10,000 in five years. If he makes annual deposits on the first of each year and the account earns 4.5% interest annually, how much should he deposit each year in order to have $10,000 in the account after the final deposit on the first of the \begin{align*}6^{th}\end{align*} year. Round your answer to the nearest $100.

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 11.10. 

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Vocabulary

induction

Induction is a method of mathematical proof typically used to establish that a given statement is true for all positive integers.

series

A series is the sum of the terms of a sequence.

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