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9.1: Exploring Geometric Sequences

Difficulty Level: At Grade Created by: CK-12
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This activity is intended to supplement Calculus, Chapter 8, Lesson 1.

The height that a ball rebounds to after repeated bounces is an example of a geometric sequence. The top of the ball appears to be about 4.0, 2.8, 2.0, and 1.4 units. If the ratios of consecutive terms of a sequence are the same then it is a geometric sequence. The common ratio \begin{align*}r\end{align*}r for these values is about 0.7.

Changing the Common Ratio

Explore what happens when the common ratio changes.

Start the Transform App. Press \begin{align*}Y=\end{align*}Y= and for \begin{align*}Y1\end{align*}Y1, enter \begin{align*}4*A ^{\land} (X-1)\end{align*}4A(X1).

Change your settings by pressing WINDOW and arrow right to go to SETTINGS. Set \begin{align*}A = 0.7\end{align*}A=0.7 and Step= 0.1.

Graph the function by pressing ZOOM and selecting ZoomStandard. Change the value of the common ratio \begin{align*}(A)\end{align*}(A).

1. What did you observe happens when you change the common ratio from positive to negative? Explain why this happens.

2. When the common ratio is larger than 1, explain what happens to the graph and values of \begin{align*}y\end{align*}y.

3. What \begin{align*}r-\end{align*}rvalues could model the heights of a ball bounce? Explain.

Changing the Initial Value and the Common Ratio

Press \begin{align*}Y=\end{align*}Y= and change \begin{align*}Y1\end{align*}Y1 to \begin{align*}B*A ^{\land} (X-1)\end{align*}BA(X1).

Change the SETTINGS so that \begin{align*}A = 0.7\end{align*}A=0.7, the \begin{align*}B = 4\end{align*}B=4 and \begin{align*}\text{Step} = 0.1\end{align*}Step=0.1.

4. Explain your observations of what happens when \begin{align*}B\end{align*}B changes. What is \begin{align*}B\end{align*}B also known as?

5. If the common ratio is less than -1, describe what occurs to the terms of the sequence.

Extension – Partial Sum Formula

The sum of a finite geometric series can be useful for calculating funds in your bank account, the depreciation of a car, or the population growth of a city.

e.g. \begin{align*}S_6 = 4 + 8 + 16 + 32 + 64 + 128\end{align*}S6=4+8+16+32+64+128

In this example the common ratio is 2, the first term is 4, and there are 6 terms.

The general formula

\begin{align*}S_n = a_1 + a_2 + a_3 + \ldots + a_{n-1} + a_n\end{align*}Sn=a1+a2+a3++an1+an

Since \begin{align*}a_n = r \cdot a_{n-1}\end{align*}an=ran1, substituting gives

\begin{align*}S_n & = a_1 + r \cdot a_1 + r^2 \cdot a_1 + r^3 \cdot a_1 + \ldots + r^{n-2} \cdot a_1 + r^{n-1} \cdot a_1 \\ r \cdot S_n & = r \cdot a_1 + r^2 \cdot a_1 + r^3 \cdot a_1 + \ldots + r^{n-1} \cdot a_1 + r^{n} \cdot a_1\end{align*}SnrSn=a1+ra1+r2a1+r3a1++rn2a1+rn1a1=ra1+r2a1+r3a1++rn1a1+rna1

Subtracting the previous two lines

\begin{align*}S_n - r \cdot S_n = a_1 - r^{n} \cdot a_1\end{align*}SnrSn=a1rna1

\begin{align*}S_n(1 - r) = a_1(1 - r^n)\end{align*}Sn(1r)=a1(1rn) So \begin{align*}S_n = a_1 \cdot \frac{1-r^n}{1-r}\end{align*}Sn=a11rn1r

Use the formula to find the sum of the following finite geometric series.

6. Find \begin{align*}S_5\end{align*}S5 for \begin{align*}a_n =6 \left (\frac{1}{3} \right )^{n-1}.=\end{align*}an=6(13)n1.=

7. \begin{align*}\frac{1}{7} + \frac{1}{7^2} + \frac{1}{7^3} + \frac{1}{7^4} + \frac{1}{7^5} + \frac{1}{7^6} =\end{align*}17+172+173+174+175+176=

8. Find \begin{align*}S_{25}\end{align*} for \begin{align*}a_n = 2(1.01)^{n - 1}\end{align*}.

9. \begin{align*}64 - 32 + 16 - 8 + 4 - 2 + 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} + \frac{1}{256} =\end{align*}

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Date Created:
Feb 23, 2012
Last Modified:
Nov 04, 2014
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