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Exercises 1 We are asked to compare the sequences shown below. We can label them to tell them apart. A) 2, 4, 6, 8, 10…B) 2, 4, 8, 16, 32… Since the way in which the terms change characterizes each type of sequence, we can analyze this to classify the sequences given. To do this, let's start by finding the difference between consecutive terms.As we can see, Sequence A has a common difference of 2. Therefore, it is an arithmetic sequence. On the other hand, we can know that Sequence B is not an arithmetic sequence, since the difference between its consecutive terms is not constant. We can now check for the ratios between consecutive terms.Notice that Sequence B has a common ratio of 2. Therefore, this is a geometric sequence. We can summarize our results as shown below.The sequence 2, 4, 6, 8, 10… is an arithmetic sequence with a common difference of 2. The sequence 2, 4, 8, 16, 32… is a geometric sequence with a common ratio of 2. | |

Exercises 2 Recall that a geometric sequence can be described by the equation shown below. an=a1rn−1 In this format an is the nth term, a1 is the initial term, and r is the common ratio. Now let's review the form of an exponential equation. y=abx Here a=0, b>0, and b=1. If we compare both equations we can see that they have the same form. Note in particular that the common ratio r of a geometric sequence plays the role of the base b of an exponential function. Geometic Sequence an=a1rn−1Exponential function y=abx Therefore, if r>0 we can write an exponential function with b=r, and the points of the sequence would overlap the graph of the function.On the other hand, if r<0 this is not possible, since by definition b cannot be negative for an exponential function. In a case like this, the graph of the sequence forms a pattern of points alternating between two different quadrants instead. | |

Exercises 3 To determine if the given sequence is geometric, we will calculate the ratios between consecutive terms.The ratio between consecutive terms is the same. Therefore, the sequence is geometric with a common ratio of 3. | |

Exercises 4 The common ratio of a geometric sequence is the ratio between any two consecutive terms.The ratio between consecutive terms is the same. Therefore, the sequence is geometric with a common ratio of 61. | |

Exercises 5 To determine if the given sequence is geometric, we will calculate the ratios between consecutive terms.The ratio between consecutive terms is the same. Therefore, the sequence is geometric with a common ratio of -8. | |

Exercises 6 To determine if the given sequence is geometric, we will calculate the ratios between consecutive terms.The ratio between consecutive terms is the same. Therefore, the sequence is geometric with a common ratio of 10. | |

Exercises 7 To determine if the given sequence is geometric, we will calculate the ratios between consecutive terms.The ratio between consecutive terms is the same. Therefore, the sequence is geometric with a common ratio of 43. | |

Exercises 8 The common ratio of a geometric sequence is the ratio between any two consecutive terms.The ratio between consecutive terms is the same. Therefore, the sequence is geometric with a common ratio of -31. | |

Exercises 9 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms.We can see above that the ratios are not the same. One of them is a division by zero which is undefined. However the differences are the same. Therefore, the sequence is a arithmetic sequence with a common difference of 8. | |

Exercises 10 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms.We can see above that the differences are not the same. The ratios are not the same too. Therefore, the sequence is neither arithmetic, nor geometric. | |

Exercises 11 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms.We can see above that the differences are not the same. The ratios are not the same too. Therefore, the sequence is a neither arithmetic, nor geometric. | |

Exercises 12 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms.We can see above that the differences are not the same. However the ratios are the same. Therefore, the sequence is geometric with a common ratio of 7. | |

Exercises 13 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms.We can see above that the differences are not the same. However the ratios are the same. Therefore, the sequence is a geometric sequence with a common ratio of 81. | |

Exercises 14 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms.We can see above that the ratios are not the same. However the differences are the same. Therefore, the sequence is a arithmetic sequence with a common difference of 7. | |

Exercises 15 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms. The four terms given in the graph are 2, 10, 50, and 250.We can see above that the differences are not the same. However the ratios are the same. Therefore, the sequence is a geometric sequence with a common ratio of 5. | |

Exercises 16 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms. But first, we have to pull the point from the graph. Point (1,4) mean that a1=4, (2,11) mean that a2=11, (3,16) mean that a3=16, and (4,19) mean that a4=19.We can see above that the differences are not the same and the ratios are the same too. Therefore, the sequence is neither arithmetic, nor geometric. | |

Exercises 17 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms. The four terms given in the plot are 120, 60, 20, and 5.We can see above that neither the ratios nor the differences are the same. Therefore, the sequence is neither geometric nor arithmetic. | |

Exercises 18 We want to identify the given sequence as arithmetic, geometric, or neither. To do so we will calculate the difference and ratio between consecutive terms. But first, we have to pull the point from the graph. Point (1,-3) mean that a1=-3, (2,6) mean that a2=6, (3,15) mean that a3=15, and (4,24) mean that a4=24.We can see above that the ratios are not the same. However the differences are the same. Therefore, the sequence is a arithmetic sequence with a common difference of 9. | |

Exercises 19 Let's start by paying close attention to the terms of the given sequence. a15a220a380a4320 We can see that the ratio between any two consecutive terms is always 4. Therefore, we have a geometric sequence with a common ratio of 4. To find the next three terms, we will multiply the last term by 4 and repeat it two times. a5a6a7===320⋅41280⋅45120⋅4===1280512020480 To graph the first seven terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 20 Let's start by paying close attention to the terms of the given sequence. a1-3a212a3-48a4192 We can see that the ratio between any two consecutive terms is always -4. Therefore, we have a geometric sequence with a common ratio of -4. To find the next three terms, we will multiply the last term by -4 and repeat it two times. a5a6a7===192⋅(-4)-768⋅(-4)3072⋅(-4)===-7683072-12288 To graph the first seven terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 21 Let's start by paying close attention to the terms of the given sequence. a181a2-27a39a4-3 We can see that the ratio between any two consecutive terms is always -31. Therefore, we have a geometric sequence with a common ratio of -31. To find the next three terms, we will multiply the last term by -31 and repeat it two times. a5a6a7===-3⋅(-31)1⋅(-31)-31⋅(-31)===1-3191 To graph the first seven terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 22 Let's start by paying close attention to the terms of the given sequence. a1-375a2-75a3-15a4-3 We can see that the ratio between any two consecutive terms is always 51. Therefore, we have a geometric sequence with a common ratio of 51. To find the next three terms, we will multiply the last term by 51 and repeat it two times. a5a6a7===-3⋅51-53⋅51-253⋅51===-53-253-1253 To graph the first seven terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 23 Let's start by paying close attention to the terms of the given sequence. a132a28a32a421 We can see that the ratio between any two consecutive terms is always 41. Therefore, we have a geometric sequence with a common ratio of 41. To find the next three terms, we will multiply the last term by 41 and repeat it two times. a5a6a7===21⋅4181⋅41321⋅41===813211281 To graph the first seven terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 24 The common ratio of a geometric sequence is the ratio between any two consecutive terms.The common ratio of our sequence is 23. To obtain the next three terms, we multiply by 23.To plot these terms, we will let the horizontal axis represent the position of the term within the sequence — this is the domain — and the vertical axis will represent the value of the terms — the range. | |

Exercises 25 Explicit equations for geometric sequences follow a specific format. an=a1(r)n−1 In this form, a1 is the first term in a given sequence, r is the common ratio from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=2. Let's observe the other terms to determine the common ratio r. 2⟶×48⟶×432⟶×4128… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1(r)n−1r=4, a1=2an=2(4)n−1 This equation can be used to find any term in the given sequence. To find a6, the 6th term in the sequence, we substitute 6 for n. an=2(4)n−1n=6a6=2(4)6−1Subtract terma6=2(4)5Calculate powera6=2⋅1024Multiplya6=2048 The 6th term in the sequence is 2048. | |

Exercises 26 We want to find the equation for the nth term and the value of the sixth term of the given geometric sequence. First we will find its explicit formula. Let's start by calculating the common ratio.We will now use the common ratio -5 and the first term 0.6 to write the formula. an=a1rn−1⇔an=0.6(-5)n−1 Finally, to calculate the value of a6 we will substitute 6 for n in our formula. an=0.6(-5)n−1n=6a6=0.6(-5)6−1 Evaluate right-hand side Subtract terma6=0.6(-5)5Calculate powera6=0.6(-3125)a(-b)=-a⋅b a6=-1875 | |

Exercises 27 Explicit equations for geometric sequences follow a specific format. an=a1(r)n−1 In this form, a1 is the first term in a given sequence, r is the common ratio from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=-81. Let's observe the other terms to determine the common ratio r. -81⟶×2-41⟶×2-21⟶×2-1… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1(r)n−1r=2, a1=-81an=-81(2)n−1 This equation can be used to find any term in the given sequence. To find a6, the 6th term in the sequence, we substitute 6 for n. an=-81(2)n−1n=6a6=-81(2)6−1Subtract terma6=-81(2)5Calculate powera6=-81⋅32b1⋅a=baa6=-832Calculate quotienta6=-4 The 6th term in the sequence is -4. | |

Exercises 28 Explicit equations for geometric sequences follow a specific format. an=a1(r)n−1 In this form, a1 is the first term in a given sequence, r is the common ratio from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=0.1. Let's observe the other terms to determine the common ratio r. 0.1⟶×90.9⟶×98.1⟶×972.9… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1(r)n−1r=9, a1=0.1an=0.1(9)n−1 This equation can be used to find any term in the given sequence. To find a6, the 6th term in the sequence, we substitute 6 for n. an=0.1(9)n−1n=6a6=0.1(9)6−1Subtract terma6=0.1(9)5Calculate powera6=0.1⋅59049Multiplya6=5904.9 The 6th term in the sequence is 5904.9. | |

Exercises 29 Explicit equations for geometric sequences follow a specific format. an=a1(r)n−1 In this form, a1 is the first term in a given sequence, r is the common ratio from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=7640. Let's observe the other terms to determine the common ratio r. 7640⟶×0.1764⟶×0.176.4⟶×0.17.64… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1(r)n−1r=0.1, a1=7640an=7640(0.1)n−1 This equation can be used to find any term in the given sequence. To find a6, the 6th term in the sequence, we substitute 6 for n. an=7640(0.1)n−1n=6a6=7640(0.1)6−1Subtract terma6=7640(0.1)5Calculate powera6=7640⋅0.00001Multiplya6=0.0764 The 6th term in the sequence is 0.0764. | |

Exercises 30 We want to find the equation for the nth term and the value of the sixth term of the given geometric sequence. First we will find its explicit formula. Let's start by calculating the common ratio using the four terms given in the table, -192, 48, -12, and 3.We will now use the common ratio -41 and the first term -192 to write the formula. an=a1rn−1⇔an=-192(-41)n−1 Finally, to calculate the value of a6 we will substitute 6 for n in our formula. an=-192(-41)n−1n=6a6=-192(-41)6−1 Evaluate right-hand side Subtract terma6=-192(-41)5Put minus sign in denominatora6=-192(-41)5(ba)m=bmama6=-192(-10241)a⋅b1=baa6=-1024-192-b-a=baa6=1024192ba=b/64a/64 a6=163 | |

Exercises 31 Explicit equations for geometric sequences follow a specific format. an=a1(r)n−1 In this form, a1 is the first term in a given sequence, r is the common ratio from one term to the next, and an is the nth term in the sequence. For this exercise, the first term is a1=0.5. Let's observe the other terms to determine the common ratio r. 0.5⟶×(-6)-3⟶×(-6)18⟶×(-6)-108… By substituting these two values into the explicit equation and simplifying, we can find the formula for this sequence. an=a1(r)n−1r=-6, a1=0.5an=0.5(-6)n−1 This equation can be used to find any term in the given sequence. To find a6, the 6th term in the sequence, we substitute 6 for n. an=0.5(-6)n−1n=6a6=0.5(-6)6−1Subtract terma6=0.5(-6)5Calculate powera6=0.5⋅-7,776Multiplya6=-3888 The 6th term in the sequence is -3888. | |

Exercises 32 We want to find the equation for the nth term and the value of the sixth term of the given geometric sequence. First we will find its explicit formula. Let's start by calculating the common ratio using the four terms given in the graph, 224, 112, 56, and 28.We will now use the common ratio 21 and the first term 224 to write the formula. an=a1rn−1⇔an=224(21)n−1 Finally, to calculate the value of a6 we will substitute 6 for n in our formula. an=224(21)n−1n=6a6=224(21)6−1 Evaluate right-hand side Subtract terma6=224(21)5(ba)m=bmama6=224(321)a⋅b1=baa6=32224Calculate quotient a6=7 | |

Exercises 33 We know the initial number of teams and the numbers of teams left after the first and second rounds. We will make a table and check if there is a common ratio. Let x and y represent the number of rounds and the number of teams left, respectively.We see that there is a common ratio of 21 between the terms. We can find the next three terms by multiplying the number of teams left after the second round by 21 for each round. Let's do it!The next three terms are 16, 8, and 4, which are the numbers of teams remain after the third, fourth and fifth rounds, respectively. | |

Exercises 34 Using the given information, we will make a table and check if there is a common ratio. Let x and y represent the number of zoom out and the area displayed on the screen, respectively.We see that there is a common ratio of 4. We want to find the area when we zoom out four times. To do so, we will multiply the area after the second zoom by 4 for each round.From the table, we see that the screen area is 24576 square units after the fourth zoom. | |

Exercises 35 By examining the pattern, we see that the common ratio is -21, not -2.Now, let's find the next three terms of the sequence.The next three terms are -21, 41, and -81. | |

Exercises 36 We see that the common ratio is 6, not -6.Since the common ratio is 6, and the first term is -2, the nth term is given by the equation below. an=a1rn−1⇔an=-2(6)n−1 | |

Exercises 37 | |

Exercises 38 | |

Exercises 39 | |

Exercises 40 | |

Exercises 41 To find the number of teams that have been eliminated after each round, we will use the table in this exercise. Recall that x is the number of rounds and y is the number of teams remain.These differences shows us the number of teams eliminated. Let's check if there is a common difference between each pair of consecutive terms.There is a common ratio of 21, and the first term is 64, so it is a geometric sequence. Therefore, the nth term of the geometric sequence is given by the equation below. an=a1rn−1⇔an=64(21)n−1 | |

Exercises 42 We know that the graphing calculator screen initially shows an area of 96 square units. The perimeter of this screen is 40 units since the width and the length of the screen are 12 and 8 units respectively.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution19674_1_910203471_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = {"Graph34":{"url":"\/mediawiki\/images\/b\/b2\/Graph34.JPG"}}; try { var b=mlg.board([-7.5,4.5,7,-4],{desktopSize:"medium",style:"usa"}); var opt1 = {dash:2,"strokeWidth":1,firstArrow:{size:6},lastArrow:{size:6,type:6}};var txt2 = b.txt(0,-2.6,"\\text{Width: } \\textcolor{darkorange}{12\\text{ units}}"); var arrow2 = b.arrow(b.node(-4.8,-1.2),b.node(-4.8,3.3),opt1); var txt4 = b.txt(-5.5,1,"\\text{Length: } \\textcolor{green}{8\\text{ units}}",{rotate:90}); var arrow2 = b.arrow(b.node(-3.3,-2),b.node(3.3,-2),opt1); var func1 = b.func("1.1*0.68^x",{strokeWidth:2.3,strokeColor:"red",xMin:-2.8,xMax:3.3,firstArrow:false,lastArrow:false}); b.image(images["Graph34"].url,[-3.7,-1.2],7.5,4.5);} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19674_1_910203471_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19674_1_910203471_l", "Solution19674_1_910203471_p", 1, code); }); } ); } window.JXQtable["Solution19674_1_910203471_l"] = true;We know the area of the screens we zoom out each time. See this exercise. Let's make a table to show the perimeters of the screens displayed on the screen after we zoom out.yWrite as a product (a⋅b)Perimeter 2(a+b) 9612⋅82(12+8)=40 38424⋅162(24+16)=80 153648⋅322(48+32)=160 Let's write the first three term of the sequence.There is a common ratio of 2 making it a geometric sequence. Therefore, the nth term of the geometric sequence is given by the equation below. Note that the first term is 40. an=a1rn−1⇔an=40(2)n−1 | |

Exercises 43 Let's start by reviewing the main characteristics of each type of sequence. An arithmetic sequence is characterized for having a common difference between pairs of consecutive terms. This means that the next term can be found by adding the common difference to the current term.Since the same happens with the y-values corresponding to equally spaced x-values in a linear function, the graph of an arithmetic sequence forms a linear pattern.On the other hand, a geometric sequence is characterized for having a common ratio between pairs of consecutive terms. This means that the next term can be found by multiplying by the common ratio to the current term.Since the same happens with the y-values corresponding to equally spaced x-values in an exponential function, if the common ratio is positive, the graph of the geometric sequence forms an exponential pattern.If the common ratio is negative, then the graph of a geometric sequence forms a pattern of points alternating between two different quadrants instead. | |

Exercises 44 We are given two consecutive terms of a sequence. …,-8,0,… If the sequence is a geometric sequence, then its nth term can be found by the formula below. an=a1(r)n−1 Notice that one of the terms is 0, so we should be able to find it using the formula. 0=a1(r)n−1 By the Zero Product Property, a1 or r should be 0. We have two cases.If a1=0, then all terms have to be zero. If r=0, then all terms, with exception to the first term, have to be zero.Note that -8 is not the first term in the sequence and is not equal to zero. This means that the sequence is not a geometric sequence. | |

Exercises 45 We will use the following facts.If a sequence is arithmetic, the difference between two consecutive terms is constant. If a sequence is geometric, the ratio between two consecutive terms is constant. Let's examine the given sequence.We see that it is an arithmetic sequence with a common difference of 0. Let's check if it is also a geometric sequence.We see that there is a common ratio of 1, meaning it is also a geometric sequence. | |

Exercises 46 | |

Exercises 47 We will start by writing the formula for the nth term of the geometric sequence. We are given that r=3. an=a1(r)n−1⇔an=a1(3)n−1 Let's find a1 using a3=81. an=a1(3)n−1n=3, a3=8181=a1(3)3−1 Solve for a1 Subtract term81=a1(3)2Calculate power81=a1(9)LHS/9=RHS/99=a1Rearrange equation a1=9 The formula below gives us the nth term of this geometric sequence. an=a1(3)n−1⇔an=9(3)n−1 Now we can find the 9th term. an=9(3)n−1n=9a9=9(3)9−1 Simplify right-hand side Subtract terma9=9(3)8Calculate powera9=9(6561)Multiply a9=59049 The 9th term of the sequence is 59049. | |

Exercises 48 Let's start by reviewing the main characteristics of each type of sequence. An arithmetic sequence is characterized for having a common difference between pairs of consecutive terms. This means that the next term can be found by adding the common difference to the current term.A geometric sequence is characterized for having a common ratio between pairs of consecutive terms. This means that the next term can be found by multiplying by the common ratio with the current term.Therefore, to obtain a sequence that is not arithmetic or geometric, we just need to choose any rule for our sequence that is not adding a constant term or multiplying by a constant term. For example, consider a sequence that starts at 0 and 1, and the next term is given by the sum of the two terms before it. 0,1, 0+1 1, 1+1 2, 1+2 3, 2+3 5,… This is a special sequence called the Fibonacci Sequence. However, note that the rule can be anything, and therefore there are infinitely many solutions satisfying the exercise's conditions. | |

Exercises 49 Let's start by clarifying what we mean by dependent and independent terms. Two terms are dependent if one can be obtained from the other following a specific rule. On the other hand, two terms are independent when they are not related at all. Now, consider the geometric sequence shown below.The sequence above has a common ratio of 2, and therefore the next term can always be obtained from its previous term by multiplying by 2. Note that we can do the same for all geometric sequences by using their corresponding common ratios. Therefore, their terms are dependent. | |

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