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| 15 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
A regular sheet of paper has a thickness of about 0.1 millimeter. Every time the paper is folded in half, its thickness doubles. Try the applet below to explore how the thickness increases as the sheet keeps getting folded in half.
As can be seen in the previous applet, each time the sheet of paper was folded in half, its thickness doubled. The values for the paper thickness after each fold can be represented by terms of a specific type of sequence called a geometric sequence.
a1>0 | a1<0 | |
---|---|---|
r>1 | Increasing 3 →×2 6 →×2 12 →×2 24 →×2 48… |
Decreasing -3 →×2 -6 →×2 -12 →×2 -24 →×2 -48… |
r=1 | Constant
3 →×1 3 →×1 3 →×1 3 →×1 3… |
Constant
-3 →×1 -3 →×1 -3 →×1 -3 →×1 -3… |
0<r<1 | Decreasing 48 →×21 24 →×21 12 →×21 6 →×21 3… |
Increasing -48 →×21 -24 →×21 -12 →×21 -6 →×21 -3… |
r<0 | Alternating
3 →×(-2) -6 →×(-2) 12 →×(-2) -24 →×(-2) 48… |
Alternating
-3 →×(-2) 6 →×(-2) -12 →×(-2) 24 →×(-2) -48… |
Like for any other sequence, the first term of a geometric sequence is denoted by a1, the second by a2, and so on. Since geometric sequences have a common ratio r, once one term is known, the next term can always be found by multiplying the known term by r.
In fact, the sequence can be found using only a1 and r, since all the subsequent terms can be found by multiplying a1 by r a specific number of times. Because of this, geometric sequences have the following general form.
a1, a1r, a1r2, a1r3, a1r4, …
Geometric sequences can be described by using a formula that uses the positions of the terms to calculate their values. This formula is called an explicit rule of the geometric sequence.
Every geometric sequence can be described by a function known as the explicit rule, whose input is the position of a term n and whose output is the term's value an. An explicit rule for a geometric sequence has the following general form.
an=a1⋅rn−1
Here, a1 is the first term of the sequence and r is the common ratio.
n | an | Using a1 and r |
---|---|---|
1 | a1 | a1⋅r0 |
2 | a2 | a1⋅r1 |
3 | a3 | a1⋅r2 |
4 | a4 | a1⋅r3 |
It can be seen that the exponent of the common ratio is always 1 less than the value of the position n. With this pattern, it is possible to write the explicit rule in the same form as the formula given at the beginning.
an=a1⋅rn−1
Jordan is studying her biology notes. She finds out that a bacterium can divide into two bacteria in a period of time of about 20 minutes. These two bacteria can then divide into two bacteria each, and so on.
Having read her notes, she is now ready for the lab practice. In a glass slide she has prepared a sample with 7 isolated bacteria.
For her practice, Jordan had to check every 20 minutes and count the number of bacteria. The results can be written as a sequence.
Since this sequence has a common ratio, it is, by definition, a geometric sequence.
Ramsha's ecology teacher asks each of the 30 students in her class to plant a seed. Then, they explain that if each student asked 3 people to do the same by tomorrow, and these 3 people did the same by the next day, the amount of planted seeds could modeled by a geometric sequence with the explicit rule an=30⋅3n−1.
n=5
Subtract term
Calculate power
Multiply
n=10
Subtract term
Calculate power
Multiply
There is a famous story about the invention of chess. When the game was presented to the king, he was so happy about it that he told the inventor to choose any payment. The inventor asked the king to put a single grain of rice on the first square, two grains on the second, four on the third and so on. The amount on the final square was the desired payment.
The king was surprised and believed that this was such a bad decision for the inventor, as the king thought this debt could be paid with no more than a bag of rice. However, when he ordered his treasurer to pay the agreed amount, it turned out that this wealthy king was not rich enough as to pay the debt. In fact, it is impossible for anyone to pay it!
n=64
Subtract term
Calculate power
1⋅a=a
Round to 3 significant digit(s)
Commutative Property of Multiplication
Multiply
am⋅an=am+n
Write in scientific notation
A criminal mastermind started a big scam. They emailed some number of people and convinced them to send money by providing suspicious information for an investment plan. The criminal told each victim that all they needed to do was to contact 5 people and ask them to send the same amount of money, and the mastermind's company would take care of the rest.
The scammer gave each victim one week to find 5 people. Then, they gave one week to the new people to repeat the process. However, because of a blunder, the police caught the criminal.
n=4
a4=2500
Subtract term
Calculate power
LHS/125=RHS/125
Rearrange equation
It has been shown how an explicit rule can describe a geometric sequence with a function that receives the term position as input and returns the term's value as output. However, a geometric sequence can also be described with a recursive relation.
A recursive rule of a geometric sequence is a pair made of a recursive equation telling how the term an is related to its preceding term an−1, and the first term of the sequence a1.
a1,an=an−1⋅r
Now it will be explained, step by step, how to write the recursive rule for a geometric sequence.
The recursive rule of a geometric sequence includes the first term of the sequence and a recursive equation.
a1,an=an−1⋅r
Calculate quotient
Cancel out common factors
Simplify quotient
aam=am−1
The diagram above shows a sequence representing the percentage of the original ratio of carbon R present in the organism. The term's value an is the percentage left of the original ratio R in the sample after n half life periods.
Now the values of the first term and the common ratio will be substituted in the general form to obtain the desired recursive rule.
n=4
Subtract term
a3=12.5
a⋅b1=ba
Calculate quotient
a1=50,(an=an−1⋅21 | ||
---|---|---|
n | Substitution | an |
n=4 | a4=a3⋅21⇓a4=12.5⋅21
|
a4=6.25 |
n=5 | a5=a4⋅21⇓a5=6.25⋅21
|
a5=3.125 |
Now, the next two terms can be added to the sequence.
Ali is curious about what would happen if he dropped a ball from the top of the building he lives in, which is about 81 meters tall. He has noticed that when the ball bounces, it reaches one third the height it was dropped from, so the ball would reach first 381=27 meters high on the first bounce. To calculate the height for the next bounces, he writes a recursive rule.
n=2
Subtract term
a1=27
a⋅b1=ba
Calculate quotient
an=an−1⋅31( | ||
---|---|---|
n | Substitution | an |
n=1 | a1=27
|
a1=27 |
n=2 | a2=a1⋅31⇓a2=27⋅31
|
a2=9 |
n=3 | a3=a2⋅31⇓a3=9⋅31
|
a3=3 |
n=5
Subtract term
Calculate power
a⋅b1=ba
Calculate quotient
Round to 2 decimal place(s)
Because the values of the terms of a geometric sequence increase by a constant factor, every geometric sequence shows an exponential relation, where the common ratio can be considered as the constant multiplier of the associated exponential function. Consider the following geometric sequence.
The sequence can be represented by a table of values, where the independent variable n represents the term's position and the dependent variable an represents the term's value.
In this case, the common ratio is 2. Therefore, all the terms of the sequence lie on the graph of an exponential curve with a constant multiplier of 2.
The relationship of n and an satisfies the definition of an exponential function. Therefore, geometric sequences are exponential functions themselves — their domain and range just happen to be discrete. In other words, the variables n and an only take specific values. This is why their graphs are series of isolated points and not a solid line.n=15
Subtract term
Calculate power
Multiply
n=20
Subtract term
Calculate power
Multiply
Find the common ratio r of the geometric sequences.
We want to find the common ratio of the given geometric sequence. The common ratio is the ratio of any two consecutive terms. Therefore, to find the common ratio, we must calculate the quotient between two consecutive terms.
As we can see, the common ratio for this sequence is 4.
Again, we want to find the ratio between consecutive terms. We can do so by calculating the quotient between two consecutive terms in the sequence.
The common ratio for this sequence is 16.
Similar to the previous parts, we calculate the common ratio by calculating the quotient between consecutive terms.
The common ratio for this sequence is 10.
Determine whether each sequence is a geometric sequence.
We are want to determine whether the given sequence is geometric or not. In order to do so, we will calculate the ratio between consecutive terms.
As we can see, the ratios are not the same. One of them is even a division by zero, which is undefined! Therefore, the sequence is not geometric.
We will investigate whether the sequence is geometric by examining the ratios of consecutive terms as we did in Part A.
Again, the ratios are not the same. Therefore, the sequence is not geometric.
Once more, we will investigate the ratios of consecutive terms.
As we can see, the ratios between all pairs of consecutive terms is 7. Therefore, there is a common ratio and the sequence is geometric.
The first term of the geometric sequence is already given as part of the recursive rule. a_1 = 2, a_n=3a_(n-1) Now that we know that a_1= 2, we can use the provided rule to find the second term in the sequence a_2. To do so, we will substitute 2 for n and simplify.
We can follow the same procedure to find the next three terms in the sequence, a_3, a_4, and a_5.
n | a_n=3a_(n-1) | a_n |
---|---|---|
1 | a_1=2 | 2 |
2 | a_2=3a_(2-1) ⇕ a_2=3 a_1 |
a_2=3( 2) ⇕ a_2= 6 |
3 | a_3=3a_(3-1) ⇕ a_3=3 a_2 |
a_3=3( 6) ⇕ a_3= 18 |
4 | a_4=3a_(4-1) ⇕ a_4=3 a_3 |
a_4=3( 18) ⇕ a_4= 54 |
5 | a_5=3a_(5-1) ⇕ a_5=3 a_4 |
a_5=3( 54) ⇕ a_5= 162 |
Therefore, the first five terms of the sequence are 2, 6, 18, 54, and 162. We could also find the consecutive terms by multiplying the previous term by 3.
Write the next three terms of each geometric sequence.
Let's start by determine the common ratio of the geometric sequence.
The common ratio is 5. To find the next three terms, we will use the fact that each term after a_1 is obtained by multiplying the previous term by the common ratio. lll a_5 = 625 ( 5) & ⇒ & a_5= 3125 a_6 = 3125 ( 5) & ⇒ & a_6=15 625 a_7 = 15 625 ( 5) & ⇒ & a_7=78 125 The next three terms are 3125, 15 625, and 78 125.
As we did before, we will start by finding the common ratio by calculating the quotient between consecutive terms.
The common ratio is 14. To find the next three terms, we multiply the previous term by 14. lll a_5 = 1/2 ( 1/4) & ⇔ & a_5=1/8 [0.5em] a_6 = 1/8 ( 1/4) & ⇔ & a_6=1/32 [0.5em] a_7 = 1/32 ( 1/4) & ⇔ & a_7=1/128 The next three terms are 18, 132, and 1128.
As in the previous parts, we start by calculating the common ratio.
The common ratio is 32. To find the next three terms, we multiply the previous term by 32. lll a_5 = 6 ( 3/2) & ⇔ & a_5=9 [0.5em] a_6 = 9 ( 3/2) & ⇔ & a_6=27/2 [0.5em] a_7 = 27/2( 3/2) & ⇔ & a_7=81/4 The next three terms are 9, 272, and 814.
Find a9 by first writing an equation for the nth term of each geometric sequence.
The explicit rule of a geometric sequence follows a specific format. a_n= a_1 r^(n-1) In this form, a_1 is the first term of the sequence, r its common ratio, and a_n the nth term in the sequence. From the sequence, we see that the first term is a_1= 2. Let's now determine the common ratio.
By substituting r= 4 and a_1= 2 into the formula, we can find the formula for this sequence. a_n= 2( 4)^(n-1) Finally, we substitute n= 9 into the formula and simplify to determine the value of the ninth term.
The 9th term in the sequence is 131 072.
As in Part A, we can immediately see that the first term is 0.1. Let's proceed by finding the common ratio between consecutive terms.
The common ratio is 3. Now we can write the equation by substituting r= 3 and a_1= 0.1 into the formula for a geometric sequence. a_n= 0.1( 3)^(n-1) Next, we substitute n= 9 into the equation and evaluate.
The ninth term is 656.1.
As in previous parts, we can identify the first term by looking at the sequence. This time the first term is 15. Let's now find the common ratio between consecutive terms.
The common ratio is 15. Now we have all the information we need to write the equation of the geometric sequence. \begin{gathered} a_{n}={\color{#0000FF}{15}}\left({\color{#FF0000}{\dfrac15}}\right)^{n-1} \end{gathered} Next, we substitute n= 9 into the equation and evaluate.
The ninth term is 378 125.
A student sets the window of a graphing calculator to show an area of 96 square units.
Let's write the areas as a sequence and calculate the quotient of the consecutive terms to see if they are the same.
Since the quotient between both pairs of consecutive terms is 4, the sequence is a geometric sequence with the first term 96 and common ratio 4. Now we can write the explicit rule for this geometric sequence. a_n= 96( 4)^(n-1) We want to find the window's area after having zoomed out five times. Notice that zooming out five times corresponds to the sixth term of the sequence. Therefore, we will substitute n=6 into the formula and evaluate.
The screen area is 98 304 square units after zooming out five times.