<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation
Our Terms of Use (click here to view) and Privacy Policy (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use and Privacy Policy.

Sums of Finite Geometric Series

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

Atoms Practice
Estimated48 minsto complete
%
Progress
Practice Sums of Finite Geometric Series
Practice
Progress
Estimated48 minsto complete
%
Practice Now
Sums of Finite Geometric Series

Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day 45?

This is a geometric series, since the difference between the exercise time on any two days is greater than the difference between any prior two days. You could just add: \begin{align*}15 + (15 \cdot 1.05) + [(15 \cdot 1.05) \cdot 1.05]...\end{align*} and so on up to 45, but that would be horribly tedious. In this lesson, you will learn how to answer a question like this will little effort.

Watch This

Khan Academy: Geometric series sum to figure out mortgage payments

Guidance

A finite geometric series is simply a geometric series with a specific number of terms. For example, consider the series: 50 + 25 + 12.5 + ....The series is geometric: the first term is 50, and the common ration is (1/2).

The sum of the first two terms is 50 + 25 = 75. We can write this as S2 = 75
The sum of the first three is 50 + 25 + 12.5 = 87.5. We can write this as S3 = 87.5

To find the value of Sn in general, we could simply add together the first n terms in a series. However, this would obviously be tedious for a large value of n. Given the regular pattern in a geometric series - every term is (1/r) of the previous term, and the nth term is an = a1rn - 1 , we can use induction to prove a formula for Sn .

The sum of the first n terms in a geometric series is \begin{align*}S_n = \frac{a_1(1 - r^n)} {1 - r}\end{align*}

For example, for the series 50 + 25 + 12.5 + ... , the sum of the first 6 terms is:

\begin{align*}S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{50 \left(1 - (\frac{1} {2})^6 \right)} {1 - \frac{1} {2}} = \frac{50 \left(1 - \frac{1} {64} \right)} {\frac{1} {2}} = \frac{50 \left(\frac{63} {64} \right)} {\frac{1} {2}} = 50 \left(\frac{63} {64} \right) \left(\frac{2} {1} \right) = 98 \frac{7} {16}\end{align*}

The figure below shows the same calculation on a TI-83/4 calculator:

We can use this formula as long as the series in question is geometric.

Example A

Find the sum of the first 10 terms of a geometric series with a1 = 3 and r = 5.

Solution:

The sum is 58,593.

\begin{align*}S_n = \frac{a_1(1 - r^n)} {1 - r} = \frac{3(1 - 5^7)} {1 - 5} = \frac{3(1 - 78,125)} {-4} = \frac{3(-78,124)} {-4} = 58,593\end{align*}
Notice that because the common ratio in this series is 5, the terms get larger and larger. This means that for increasing values of n the sums will also get larger and larger. In contrast, in the series with common ratio (1/2), the terms gets smaller and smaller. This situation implies something important about the sum.

Example B

Find the sum of each series:

a) The first term of a geometric series is 4, and the common ratio is 3. Find S8.
b) The first term of a geometric series is 80, and the common ratio is (1/4). Find S7.

Solution:

a) \begin{align*}S_{8}=\frac {4(1-3^8)}{1-3}=13,120\end{align*}
b) \begin{align*}S_{7}= \frac {80 \left ( {1- \left ( \frac{1}{4} \right )^7} \right )}{1-\frac {1}{4}} \approx 106.66 \end{align*}

Example C

Prove the formula \begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*} by induction.

Solution:

1. If n = 1, the nth sum is the first sum, or a1 . Using the hypothesized equation, we get \begin{align*}S_1 = \frac{a_1(1 - r^1)} {1 - r} = \frac{a_1(1 - r)} {1 - r} = a_1\end{align*}. This establishes the base case.
2. Assume that the sum of the first k terms in a geometric series is \begin{align*}S_k = \frac{a_1(1 - r^k)} {1 - r}\end{align*}.
3. Show that the sum of the first k+1 terms in a geometric series is \begin{align*}S_{k + 1} = \frac{a_1(1 - r^{k + 1})} {1 - r}\end{align*}.
\begin{align*}S_{k + 1} = S_k + a_{k + 1}\end{align*} The \begin{align*}k + 1\end{align*} sum is the \begin{align*}k^{th}\end{align*} sum, plus the \begin{align*}k + 1\end{align*} term
\begin{align*} = \frac{a_1(1 - r^k)} {1 - r} + a_1r^{k + 1 - 1}\end{align*} Substitute from step 2, and substitute the \begin{align*}k + 1\end{align*} term
\begin{align*}= \frac{a_1(1 - r^k)} {1 - r} + \frac{a_1r^k(1 - r)} {1- r}\end{align*} The common denominator is \begin{align*}1 - r\end{align*}
\begin{align*}= \frac{a_1(1 - r^k) + a_1r^k(1 - r)} {1 - r}\end{align*} Simplify the fraction
\begin{align*}= \frac{a_1 \left[1 - r^k + r^k(1 - r) \right]} {1 - r}\end{align*}
\begin{align*}= \frac{a_1 \left[1 - r^k + r^k - r^{k + 1} \right]} {1 - r}\end{align*}
\begin{align*}= \frac{a_1 \left[1 - r^{k + 1} \right]} {1 - r}\end{align*} It is proven.

Therefore we have shown that \begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*} for a geometric series. Now we can use this equation to find any sum of a geometric series.

Concept question wrap-up:

"Anna is on a progressive workout plan, every day she adds 5% to her exercise time. If she starts by exercising 15mins on the first day, how many minutes will she have exercised all together on day 45?"

Use the formula: \begin{align*}S_{n}=\frac{a_1(1 - r^n)} {1 - r}\end{align*}

\begin{align*}S_{n} = \frac{15 (1 - 1.05^{45})}{1 - 1.05}\end{align*}
\begin{align*}S_{n} = 2395.5\end{align*} minutes.

-->

Guided Practice

1) Find the sum: \begin{align*}5 + 10 + 20 +...+ 640\end{align*} (Hint: if an = 640 , what is n?)

2) Use a geometric series to answer the question:

In January, a company’s sales totaled $11,000. It is predicted that the company’s sales will increase 5% each month for the next year. At this rate, what will be the total sales for the year?

3) Write the first 5 terms of the sequence: \begin{align*}-5 \cdot \frac{3}{4}^n\end{align*}

4) Write the 3rd, 4th, and 6th terms of: \begin{align*}(3)^{\left(\frac{n}{2}\right)}\end{align*}

5) Find the sum of the series: \begin{align*}\sum_{n = 1}^6 \left(-\frac{3}{2}\right)^{n-1}\end{align*}

Answers

1) \begin{align*}S_{8}=\frac {5(1-2^8)}{1-2} = 1275\end{align*}

2) \begin{align*}S_{12}=\frac {11000(1-1.05^{12})}{1-1.05} = \$175,088.39\end{align*}

3) Just do the multiplication for each term \begin{align*}n = 0 \to n = 4\end{align*}

\begin{align*}-5 \cdot \frac{3}{4}^0 \to -5 \cdot 1 \to -5\end{align*} ..... for \begin{align*}n = 0\end{align*}
\begin{align*}-5 \cdot \frac{3}{4}^1 \to -5 \cdot \frac{3}{4} \to -\frac{15}{4} \to -3.75\end{align*} ..... for \begin{align*}n = 1\end{align*}
\begin{align*}-5 \cdot \frac{3}{4}^2 \to -5 \cdot \frac{9}{16} \to -\frac{45}{16} \to -2.8\end{align*} ..... for \begin{align*}n = 2\end{align*}
\begin{align*}-5 \cdot \frac{3}{4}^3 \to -5 \cdot \frac{27}{64} \to -\frac{135}{64} \to -2.1\end{align*} ..... for \begin{align*}n = 3\end{align*}
\begin{align*}-5 \cdot \frac{3}{4}^4 \to -5 \cdot \frac{81}{256} \to -\frac{405}{256} \to -1.6\end{align*} ..... for \begin{align*}n = 4\end{align*}
\begin{align*}\therefore\end{align*} the first 5 terms are: \begin{align*}-5, -3.75, -2.8, -2.1, -1.6\end{align*}

4) As with problem 3, just perform the operations on the indicated values of n:

\begin{align*}3^{\frac{3}{2}} \to \sqrt{3^3} \to \sqrt{27} \to 5.2\end{align*} ..... for \begin{align*}n = 3\end{align*}
\begin{align*}3^{\frac{4}{2}} \to 9\end{align*} ..... for \begin{align*}n = 4\end{align*}
\begin{align*}3^{\frac{6}{2}} \to 3^3 \to 27\end{align*} ..... for \begin{align*}n = 6\end{align*}
\begin{align*}\therefore\end{align*} the 3rd, 4th, and 6th terms are: \begin{align*}5.2, 9, 27\end{align*}

5) To find the sum of the series \begin{align*}\sum_{n = 1}^6 \left(-\frac{3}{2}\right)^{n-1}\end{align*}

We could calculate all of the values for \begin{align*}n = 1 \to 6\end{align*}and add them, getting:
\begin{align*}1+ \frac{-3}{2} + \frac{9}{4} + \frac{-27}{8} + \frac{81}{16} + \frac{-243}{32} = \frac{-133}{32}\end{align*}
Or we can use the formula: \begin{align*}\left( \frac{1 - r^k}{1 - r} \right)\end{align*}
\begin{align*}\left( \frac{1 - \left(\frac{-3}{2}\right)^6}{1 - \left(\frac{-3}{2}\right)} \right) = \frac{-133}{32}\end{align*}

Explore More

Find the sum of the finite series. You may simply calculate the individual terms and add them, or you may use the formula: \begin{align*}S_n = \frac{a_1 (1- r^n)}{1- r}\end{align*}

  1. \begin{align*}1 + \left(-\frac{1}{2}\right) + \frac{1}{4} + ... + \frac{1}{64}\end{align*}
  2. \begin{align*}-6 + 12 -24 + ... -6144\end{align*}
  3. \begin{align*}(-4) + (-12) + (-36) + ... + (-2916)\end{align*}
  4. \begin{align*}9 + (-45) + 225 + ... + (-17,578,125)\end{align*}
  5. \begin{align*}\sum_{n=1}^5 -(2)^{n-1}\end{align*}
  6. \begin{align*}\sum_{n=1}^{10} 6\left(\frac{1}{2}\right)^{n-1}\end{align*}
  7. \begin{align*}\sum_{n=1}^6 8 \cdot 3^{n-1}\end{align*}
  8. \begin{align*}(-5) + 10 + (-20) + ... + (-1280)\end{align*}
  9. \begin{align*}(-9) + \frac{9}{2} + \left(-\frac{9}{4}\right) + ... + \left(-\frac{9}{16}\right)\end{align*}
  10. \begin{align*}\sum_{n=1}^9 4 \cdot \left(\frac{1}{2}\right)^{n-1}\end{align*}
  11. \begin{align*}\sum_{n=1}^{10} 4 \cdot \left(-\frac{2}{3}\right)^{n-1}\end{align*}
  12. \begin{align*}(-3) + \left(-\frac{3}{2}\right) + \left(-\frac{3}{4}\right) + ... + \left(-\frac{3}{1024}\right)\end{align*}
  13. \begin{align*}\sum_{n=1}^{11} -9 \cdot (-2)^{(n-1)}\end{align*}
  14. \begin{align*}\sum_{n=1}^9 -9 \cdot \left(-\frac{5}{3}\right)^{n-1}\end{align*}
  15. \begin{align*}\sum_{n=1}^6 -2 \cdot \left(-\frac{5}{4}\right)^{n-1}\end{align*}
  16. \begin{align*}\sum_{n=1}^{11} 8 \cdot \left(-\frac{1}{3}\right)^{n-1}\end{align*}

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 7.9. 

Vocabulary

finite series

A series is finite if it has a defined ending value.

geometric series

A geometric series is a geometric sequence written as an uncalculated sum of terms.

induction

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

infinite series

An infinite series is the sum of the terms in a sequence that has an infinite number of terms.

series

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

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Sums of Finite Geometric Series.
Please wait...
Please wait...