## Carnegie Learning Algebra I Student Text Volume 1 3rd Edition Chapter 4 Sequences

**Page 214 Problem 1 Answer**

In the question, we have been given the four figures and we are supposed to analyze the figure with respect to the number of dots given in the figure.

For this, we will first calculate the total number of dots in each figure and try to formulate the same.

The first figure has 25 dots, the second has 21 dots, the third has 17 dots and the fourth has 13 dots.

We can formulate the sequence as-

Figure1=6×4+1

Figure2=5×4+1

Figure3=4×4+1

Figure4=3×4+1

Hence, in general, we can write as Figure n=4n+1 where n is the figure sequence number.

The number of dots in the given figures are 25,21,17,13.

The given pattern can be generalized as Figure n=4n+1 where n is the figure sequence

**Page 214 Problem 2 Answer**

We have been given a figure that consists of four dots patterns.

Our task is to draw the next three figures of the pattern.

Using the formula from part (a), we will find the result.

We have the formula 4n+1, where n represent the figure sequence number.

Now we need to find the next three patterns.

Implies the sequence number will be 2,1,0.

Then we get the number of dots as:

4(2)+1=9

4(1)+1=5

4(0)+1

Now we will draw the patterns, shown below.

The next three-figure of the pattern will be:

**Page 214 Problem 3 Answer**

We have been given a figure that consists of four dots patterns.

We have to write the sequence numerically to represent the number of dots in each of the first 7 figures.

Using the part (a),(b), we will write the sequence numerically.

From the part (a),(b), we have the sequence of the number of dots in the below figures as 25,21,17,13,9,5,1.

So, the numerical sequence is 25,21,17,13,9,5,1.

Here the sequence numerically represents the number of dots in each of the first 7 figures is 25,21,17,13,9,5,1

**Page 215 Problem 4 Answer**

We have been given a figure that consists of a pattern of the number of small squares.

We need to analyze the number of small squares in each figure and also describe the pattern.

First, we will first calculate the total number of small squares in each figure and then we will formulate it.

Here we have the figure shown below.

From the figure, we observe the following,

The first big square consists of 49 small square as it has seven rows and seven columns.

The second big square consists of 36 small squares as it has six rows and six columns.

The third big square consists of 25 small squares as it has five rows and five columns.

The fourth big square consists of 16 small squares as it has four rows and four columns.

Now in general, we can formate the number of the small squares as n^{2}, where n is the figure sequence number.

The pattern of the number of the small square in each figure is 49,36,25,16 and the pattern will be given by n^{2}.

**Page 215 Problem 5 Answer**

We have been given a figure that consists of a pattern of the number of small squares.

We need to draw the next three figures of the pattern.

Using the pattern from the part (a), we will find the result.

From the part (a), we have the pattern as n^{2} where n is the number of the figure sequence.

Then we will have the next third sequence as:

32=9

22=4

12=1

The next three figures of the pattern will be:

**Page 215 Problem 6 Answer**

We have been given a figure that consists of a pattern of the number of small squares.

We need to write the sequence numerically to represent the number of small squares in each of the first 7

Figures Using the part (a),(b), we will find the result.

From the part (a),(b), we have a numeric sequence to represent the number of small squares in each of the first 7 figures will be:

72=49

62=36

52=25

42=16

Also, we will have:

32=9

22=4

12=1

So, we get the numerical sequence as 49,36,25,16,9.4.1.

The numeric sequence of the number of the small square in each figure is 49,36,25,16,9,4,1

**Page 215 Problem 7 Answer**

We are given that AI begins with 150 eggs to make omelets.

After making 1,2and 3 omelets he has left 144,138,132 eggs respectively.

We are required to determine the number of eggs AI has left after making each omelets.

Here, we will solve this by using the definition of Arithmetic Sequence.

AI begins with=150 eggs

After making one omelet,the number of eggs left with him =150−6

=144

After making 2 omelets,the number of eggs left with him=144−6

=138

And finally,

After making 3 omelets the number of eggs left with him =138−6

=132

We can observe that in each case we subtract the number six from the previous answer to get the next term.

The number of eggs AI has left after making each omelets is 150,144,138,132.

Here, we can observe the pattern that in each case we subtract the number six from the previous answer to get the next term.

**Page 215 Problem 8 Answer**

We are given that AI begins with 150 eggs to make his famous omelets, After making one omelets he has left 144 eggs.

After making 2 omelets he has 138 eggs left. After making 3 omelets he has 132 eggs left.

We are required to determine the number of eggs left after AI makes the next two omelets.

Here, we will solve this by using the definition of Arithmetic Sequence.

AI begins making of omelets with =150

After making one omelets the number of eggs left with him=150−6=144

After making 2 omelets, the number of eggs left with him=144−6=138

After making 3 omelets, the number of eggs left with him =138−6 =132

Hence, continuing, we get, After making 4 omelets, the number of eggs left with him =132−6=126

After making 5 omelets, the number of eggs left with him=126−6=120

The number of eggs left after AI makes the next two omelets are 126,120, respectively, which is obtained using the common difference.

**Page 215 Problem 9 Answer**

We are given that Al begins with 150 eggs to make his famous omelets.

After making one omelets,he has 144 eggs left.

After making 2 omelets, he has 138 eggs left. After making 3 omelets, he has 132 eggs left.

We are required to determine the sequence of the number of eggs left after AI makes each of the first five omelets.

Here, we will solve this by using the definition of Arithmetic Sequence.

We calculated the number of the number of eggs left after Al makes each of the first 5 omelets, and got the answers, 150,144,138,132,126,120, which is the required sequence of numbers. (refer previous exercise)

The sequence of number of eggs left after AI makes each of the first five omelets is150,144,138,132,126,120, obtained using the common difference 6.

**Page 216 Problem 10 Answer**

We are given that Mario begins the mosaic with a single tile. Then he adds to single tile to create second square made up of 4 tiles.

The third square he adds is made up of 9 tiles, and four square he adds is made up of 16 tiles.

We are required to determine the number of tiles in each square and describe the pattern.

Here, we will use the definition of perfect square.

The number of tile in first square is,1{2}=1

The number of tiles in second square is,2{2}=4

The number of tiles in third square is,3{2}=9

The number of tiles in fourth square is, 4{2}=16

We can observe that each terms represent consecutive perfect square terms.

The number of tiles in each square is1,4,9,16, and we can observe the pattern present is that they are consecutive perfect squares.

**Page 216 Problem 11 Answer**

We are given that first square is made up of single tile, 2nd square is made up of four tiles, 3rd square is made up of nine tiles,4th square is made up of 16 tiles.

We are required to determine the number of tiles in the next two squares.

Here, we have to find out next two squares i.e. 5th and 6th.

In first square , number of tiles is, 1×1=1

In 2^{nd} square , number of tiles is, 2×2=4

In 3^{rd} square, no. of tiles is,3×3=9

In 4^{th} square, no. of tiles is,4×4=16

In 5^{th} square, the no. of tiles is,5×5=25

In 6^{th} square, the no. of tiles is,6×6=36

The number of tiles in next two squares are 25,36, which is obtained by finding consecutive perfect squares.

**Page 216 Problem 12 Answer**

We are given that Mario begins the mosaic with a single tile.

Then he adds to single tile to create second square made up of 4 tiles.

The third square he adds is made up of 9 tiles, and four square he adds is made up of 16.

We are required to determine the sequence of number of tiles in each of the first six square.

Here, we will solve this by using the definition of Perfect squares.

We calculated the number of tiles in each of the 6 squares got the answers, 1,4,9,16,25,36, which is the required sequence of numbers. (refer previous exercise) First mario used one tile, which is the square of 1.

Then mario used 4 tiles ,which is the square of 2.Next he used 9 tiles which is the square of 3.

Next used 16 tiles which is the square of 4.In the fifth time he will use 25 tiles ,which is the square of 5.

For the sixth time he will use 36 tiles which is the square of 6.

Hence from

12=1,22

=4,32

=9,42

=16,52

=25,62

=36

we get The Final Answer,The sequence of number of tiles in each of the first six square is 1,4,9,16,25,36, which is obtained by finding the consecutive perfect squares.

**Page 217 Problem 13 Answer**

We have given a sequence of polygons that has one more side than the previous polygon.

We need to analyze the number of sides in each polygon and describe the pattern.

We are going to use the concepts of arithmetic progression to solve the question.

We have:

Number of sides in the nth term

Therefore,a_{1}=3,a_{2}=4,a_{3}=5 and a_{6}=6

Now,d=a_{2}−a_{1}

d=4−3

∴d=1

Again,d=a_{3}−a_{2}

d=5−4

∴d=1

Since the value of d is constant.

Hence, the given sequence 3,4,5,6 is an arithmetic progression.

The pattern of the polygon given in the figure, i.e., 3,4,5,6,.. is an arithmetic progression.

**Page 217 Problem 14 Answer**

In the following question, we have been given a figure.

We need to draw the next two figures of the pattern With the help of the given figure, we will find the result.

In the given figure, we get

Number of sides of the first figure =3

Number of sides of the second figure =4

The number of sides of the third figure =5

number of sides of the fourth figure =6

The number of sides of the fifth figure =7

The number of sides of the sixth figure =8

Then the next two figures will be:

The next two figures of the pattern will be:

**Page 217 Problem 15 Answer**

We have given a sequence of different types of polygons.

We need to find the sequence of the number of sides first six polygons.

We are going to use the concepts of arithmetic progression to solve the question.

We have:a_{1}=3,a_{2}=4,a_{3}=5 and a_{4}=6

Let the common difference bed,

d=a{2}−a{1}

=4−3

∴d=1

d=a{3}−a{4}

=5−4

∴d=1

Since the value of d is constant.

Therefore, the given sequence, i.e.,3,4,5,…is an arithmetic progression.

Now, we will be finding the 5^{th} term and 6^{th} term,

a_{5}=3+(5−1)1=3+4

∴a_{5}=7

a_{+}=3+(6−1)×1

=3+5

∴a_{6}=8

The sequence of the number of sides of the first six polygons is 3,4,5,6,7,8, which is in an arithmetic progression.

**Page 218 Problem 16 Answer**

In this question, we have been given Jacob’s pizza has a_{6}-foot diameter, he starts to cut one whole pizza in half.

After that he cuts each of those slices in half, then he cuts each of those slices in half, and so on.

We need to think about the size of each slice concerning the whole pizza and describe the pattern.

By using the patterns, we will calculate the result.

First, we find out the common ratio

r=a_{2}/a_{1}

=2/1

=2

Similarly, we can find out r=a_{3}/a_{2}

The 4^{th} term is =1×24−1

=23

=8

The 5^{th} term is =1×25−1

=24

=16

The pattern of the size of each slice concerning whole pizza is1,2,4,8,16. Since each contestant not only has to bake a delicious pizza, but they have to make the largest pizza they can.

**Page 218 Problem 17 Answer**

In this question, we have been given Jacob’s pizza has a 6-foot diameter, he starts to cut one whole pizza in half.

After that he cuts each of those slices in half, then he cuts each of those slices in half, and so on.

We need to find out the size of each slice compared to the original after the next two cuts.

By using the geometric sequence, we will calculate the result.

Here, the common difference is

r=a_{2}/a_{1}

=2/1

=2

The 4^{th} term is a4

=1×24−1

=1×23

=8

Then the5^{th} term is a5

=1×25−1

=1×24

=16

The size of each slice compared to the original after the next two cuts are 8 and 16.

Since each contestant not only has to bake a delicious pizza, but they have to make the largest pizza they can.

**Page 218 Problem 18 Answer**

In this question, we have been given Jacob’s pizza has a 6-foot diameter, he starts to cut one whole pizza in half.

After that he cuts each of those slices in half, then he cuts each of those slices in half, and so on.

We need to write the sequence numerically to represent the size of each slice compared to the original after each of the first 5 cuts.

By using the geometric sequence, we will calculate the result.

Here, the common ratio is

r=a_{2}/a_{1}

=2/1

=2

Now, by putting the formula the 4^{th} term is

=1×24−1

=23

=8

the5^{th} term is

=1×25−1

=24

=16

The sequence of the size of each slice compared to the original after each of the first five cuts is 1,2,4,8,16. Since each contestant not only has to bake a delicious pizza, but they have to make the largest pizza they can.

**Page 218 Problem 19 Answer**

In this question, we have been given Miranda’s uncle purchased a rare coin for $5.

He claims that the value of the coin will triple each year, so the value of the coin will be $15 next year, again the value of the coin will be $45 and $135 in 2 years and 3 years respectively.

We need to find out how the coin value changes each year i.e. how many times the value of the coin increased according to the previous year.

By using the geometric sequence, we will calculate the result.

The common ratio in this sequence is

r=a_{2}/a_{1}

=15/5

=3

We can find the next two by putting the formula,

The 5^{th} term is

=5×35−1

=5×34

=405

The 6^{th} term is

=5×36−1

=5×35

=1215

The value of the coin increased 3 times according to the previous year and the pattern is 5,15,45,135,405,1215.

**Page 218 Problem 20 Answer**

In this question, we have been given Miranda’s uncle purchased the rare coin for $5.

He claims that the value of the coin will triple each year.

So the value of the coin in the first year will become $15, then in 2 years, the value of the coin will become $45, in 3 years the value of the coin will become $135.

We need to find out the value of the coin after 4 years and 5 years i.e. next two-term a5,a6.

By using the geometric sequence, we will calculate the result.

Here, the common ratio is

r=a_{2}/a_{1}=3

Now, we can see that the next term is

The 5^{th} term is

=5×35−1

=5×34

=405

The 6^{th} term is

=5×36−1

=5×35

=1215

The value of the coin after 4 years and after 5 years are 405,1215 respectively.

Since he claims that the value of the coin will triple each year.

**Page 218 Problem 21 Answer**

In this question, we have been given Miranda’s uncle purchased the rare coin for $5 He claims that the value of the coin will triple each year.

So the value of the coin in the first year will be $15, then in two years, the value of the coin will be $45, then in three years, the value of the coin will be$135

We need to find the sequence of the value of the coin after each of the first five years. By using the geometric sequence, we will calculate the result.

Here the common ratio is

r=a_{2}/a_{1}

=15/5

=3

Now, we can find the 5^{th} term is

a_{5}=5×35−1

a_{5}=5×34

a_{5}=405

The 6^{th} term is a_{6}=5×36−1

a_{6}=5×35

a_{6}=1215

The sequence of the value of the coin after each of the first five years is 5,15,45,135,405,1215.

Since he claims that the value of the coin will triple each year.

**Page 220 Problem 22 Answer**

We have been given a sequence in which has first term is 64, and after that, each term is calculated by dividing the previous term by 4.

We have been asked that If Margaret is correct, explain why, and If Jasmine is correct, predict the next two terms of the sequence.

We will use dividing and calculate the result.

Jasmine is accurate as it does not stipulate that the sequence cannot include fractions.

Margaret would be correct if the sequence reflected anything that could only be expressed by full numbers.

Now, we are finding the next two-term,5th term=a×r5−1

=64×(1/4)4

=0.25

Further solving, 6^{th} term=a×r6−1

=64×(1/4)5

=0.0625

Jasmine is accurate as the sequence can include fractions and Margaret would be correct if the series represented only whole integers and the next two terms of this sequence are 0.25,0.0625.

**Page 221 Problem 23 Answer**

We have been given a pattern. We have been asked to explain the pattern shown in the figure is finite or infinite.

Here the pattern represents a finite sequence.

Each figure must have a natural number of blocks, hence the sequence must terminate with the last figure with only one block.

As a result, the sequence is complete.

There is no way to represent the next term with a figure because numerically, the next term is 0.

The pattern represents a finite sequence there is no way to represent the next term with a figure because numerically, the next term 0.