Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions Exercise 4.6

Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 4 Writing Linear Functions

Page 121 Essential Question  Answer

Given: Patten is an arithmetic sequence

To find a Method to describe the pattern using an arithmetic sequence.
An arithmetic sequence is an ordered list of numbers in which the difference between each pair of consecutive terms, or numbers in the list, is the same.

To describe the pattern using an arithmetic sequence, take the difference of their consecutive term and you get these all differences are same.

 

Page 122  Exercise 2  Answer

Given: Pattern is an arithmetic sequence.

To find An example from real life to describe an arithmetic sequence.

Real-life example: In a passenger train, there are 125 passengers in the first carriage, 150 passengers in the second carriage and 175 passengers in the third carriage. If the total carriage in a train is 8, then find the number of passengers in 6th
carriage.

So, in this real-life problem, the number of passengers in the carriage is in arithmetic sequence because there is a common difference(d) between The number of passengers i.e.​d=150−125
d=25 and first term(a) is 125.

Real-life example: In a passenger train, there are 125 passengers in the first carriage, 150 passengers in the second carriage, and 175 passengers in the third carriage. If total carriage in a train is 8, then find the number of passengers in 6^thcarriage.

 

Page 122  Exercise 3  Answer

Given :

To find the Number of atoms in 23 molecules.

In order to find the solution, count the number of atoms till n=5
and make a sequence then find a₂₃.

Count the number of atoms in their respective number of molecules.
So, the sequence will be, 3,6,9,12,15,…….

Here, ​a₁=3
d=6−3
=3

For a₂₃, put n=23 in a_n=a₁+(n−1)d
a₂₃=a1+(23−1)d
a₂₃=3+22×3
a₂₃=69
​
There are 69 number of atoms in 23 molecules.

 

Page 124  Exercise 1  Answer

Given sequence: 1,8,15,22,…

To find the next three terms of the arithmetic sequence.
In order to find the solution, get the value of first term a1 and common difference d then use the formula
a_n=a₁+(n−1)d

As the given sequence is 1,8,15,22,…
And ​a₁=1
d=8−1
=7

For the next three terms, calculate a₅,a₆, and a₇.
For a₅, put n=5 in above formula
​a_n=a₁+(5−1)d
a₅=(1)+(4)⋅(7)
=1+28
=29
​
For a₆, put n=6
​a_n=a₁+(6−1)d
a₆=(1)+(5).(7)
=1+35
=36
​

For a₇,put n=7
​a_n=a₁+(7−1)d
a₇=(1)+(6).(7)
=1+42
=43
​
The next three terms of the arithmetic sequence will be 29,36,43.

 

Page 124  Exercise 2  Answer

Given sequence: 20,14,8,2,…

To find the next three terms of the arithmetic sequence.
In order to find the solution, get the value of the first term a₁ and common differences then use the formula
a_n=a₁+(n−1)d

As the given sequence is 20,14,8,2,…
And ​a₁=20
d=14−20
=−6

For the next three terms, calculate a₅, a₆, and a₇.
For a₅, put n=5 in  above formula
​a_n​=a₁+(5−1)d
a₅=(20)+(4).(−6)
=20−24
=−4
​

For a₆ , put n=6 in above formula
​a_n​= a₁+(6−1)d
a₆= (20)+(5)⋅(−6)
= 20−30
= −10
​

For a₇, put n=7
​a_n​= a₁+(7−1)d
a₇= (20)+(6)⋅(−6)
= 20−36
=−16
So, the next three terms will be−4,−10,−16.

The next three terms of the arithmetic sequence will be −4,−10,−16.

 

Page 124  Exercise 3  Answer

Given sequence: 12,21,30,39,…

To find the next three terms of the arithmetic sequence.
In order to find the solution, get the value of the first term, a_1, and common difference,d then use the formula
a_n= a₁+(n−1)d

As the given sequence is 12,21,30,39,…
And ​a₁=12
d=21−12
​=9

For the next three terms, calculate a₅, a₆, and  a₇
For a₅, put n=5 in above formula
​a_n= a₁+ (5−1)d
a= (12)+ (4)⋅(9)
= 12 + 36
= 48
​

For a₇, put n=6
a_n= a₁ + (6−1)d
a₇= (12) + (5).(9)
= 12 + 45
= 57
​

For a₇, put n=7
a_n ​= a₁ + (7−1)d
a₇ = (12) + (6)⋅(9)
= 12 + 54
= 66
​
The next three terms of the arithmetic sequence will be 48,57,66.

 

Page 124   Exercise 4   Answer

Given sequence: 5,12,19,26,…

To find the Next three terms of the arithmetic sequence.
In order to find the solution, get the value of the first term,a_1, and common difference,d then use the formula
a_n=a₁+(n−1)d

As the given sequence is 5,12,19,26,…
And ​a₁=5
d=12−5
​=7

For the next three terms, calculate a₅, a₆, and a₇
For a₅, put n=5 in above formula
​a_n=a₁+(5−1)d
a₅=(5)+(4).(7)
=5+28
=33
​

For a₆, put n=6 in above formula
​a_n=a₁+(6−1)d
a₆=(5)+(5)⋅(7)
=5+35
=40
​

For a₇ , put n=7
a_n=a₁+(7−1)d
a₇=(5)+(6)⋅(7)
=5+42
=47

​So, the next three terms will be 33,40,47.
The next three terms of the arithmetic sequence will be 33,40,47.

 

Page 124 Exercise 5 Answer

Given sequence: 3,7,11,15,…

To find the next three terms of the arithmetic sequence.
In order to find the solution, get the value of the first term, a_1, and common difference,d then use the formula
a_n=a₁+(n−1)d

As the given sequence is 3,7,11,15,…
And ​a₁=3
d=7−3
=4

For the next three terms, calculate a₅,a₆, and a₇.
For a5, putn=5 in above formula
​a_n​=a₁+(5−1)d
a₅=(3)+(4).(4)
=3+16
=19
​
For a₆, put n=6
​a_n=a₁+(6−1)d
a₆=(3)+(5)⋅(4)
=3+20
=23
​
For a₇, put n=7
a_n=a₁+(7−1)d
a₇=(3)+(6)⋅(4)
=3+24
=27

​So, the next three terms will be 19,23,27.
The next three terms of the arithmetic sequence will be 19,23,27.

 

Page 124 Exercise 6 Answer

Given sequence: 2,14,26,38,…

To find the next three terms of the arithmetic sequence.

In order to find the solution, get the value of the first term,a_1, and common difference,d then use the formula
a_n=a₁+(n−1)d

As the given sequence is2,14,26,38,…
And ​a₁=2
d=14−2
=12

For the next three terms, calculate a₅,a₆, and a₇
For a₅, putn=5 in above formula
​an​=a₁+(5−1)d
a₅=(2)+(4).(12)
=2+48
=50
​

For a₆, put n=6
​a_n=a1+(6−1)d
a₆=(2)+(5)⋅(12)
=2+60
=62
​
For a₇, put n=7
​a_n​=a1 +(7−1)d
a₇=(2)+(6)⋅(12)
=2+72
=74

​So, the next three terms will be 50,62,74.
The next three terms of the arithmetic sequence will be 50,62,74.

 

Page 124 Exercise 9 Answer

Given sequence is : \(\frac{15}{2}, \frac{13}{2}, \frac{11}{2}, \frac{9}{2}\),……

To find a Graph the given arithmetic sequence.
In order to find the solution, use the order of terms,n in x-axis and the value of terms in y-axis.

To make a graph, plot the points : \(\left(1, \frac{15}{2}\right)_t\left(2, \frac{13}{2}\right)_1\left(3, \frac{11}{2}\right)_{,}\left(4, \frac{9}{2}\right)\).

Here, the order of terms is in the x-axis, and the value of terms in the y-axis.The graph will be

 

Graph of a given arithmetic sequence

 

Page 124  Exercise 10  Answer

Given sequence is 1,2.5,4,5.5,…

To find a Graph of the given arithmetic sequence.
In order to find the solution, use the order of terms,n in x -axis, and the value of terms in y-axis.

To make a graph, plot the points :(1,1),(2,2.5),(3,4),(4,5.5).
Here, the order of terms is on x-axis, and the value of terms on y-axis.

The graph will be

 

 

Graph of a given arithmetic sequence

 

Page 124 Exercise 12 Answer

Given sequence is: \(\frac{1}{4}, \frac{5}{4}, \frac{9}{4}, \frac{13}{4}\),…….

To find Graph the given arithmetic sequence.
In order to find the solution, use the order of terms,n in x- axis and the value of terms in y-axis.

To make graph , plot the points :\(\left(1, \frac{1}{4}\right)_1,\left(2, \frac{5}{4}\right)_{,}\left(3, \frac{9}{4}\right)_1\left(4, \frac{13}{4}\right)\).

Here, the Order of terms is in x-axis and the value of terms in y-axis. The Graph will be

 

 

Graph of given arithmetic sequence

 

Page 125 Exercise 13 Answer

Given: Graph is

To find whether the representation is an arithmetic progression or not.
First, read the graph and write the sequence then check whether the difference between any two consecutive terms is the same or not.

By observing the graph, the ordered pairs are such that the first coordinate is the position of the term and the second coordinate is the value of the term.
So, the sequence will be a₁ =0,a₂=1,a₃=4,a₄=9

Checking whether the sequence is arithmetic progression or not.
Calculating the difference of all consecutive terms.

​a₂−a₁
=1−0
=1

a₃−a₂
=4−1
=3

As the differences are not the same. So, the sequence is not an arithmetic progression.
The graph does not represent a_n arithmetic progression given by 0,1,4,9,…

 

Page 125 Exercise 14 Answer

Given: Graph is

To find whether the representation is an arithmetic progression or not.
First, read the graph and write the sequence then check whether the difference between any two consecutive terms is the same or not.

By observing the graph, the ordered pairs are such that the first coordinate is the position of the term and the second coordinate is the value of the term.
So, the sequence will be a₁=20, a₂=30, a₃=40, a₄=50

Checking whether the sequence is arithmetic progression or not.
Calculating the difference of all consecutive terms.
​a₂−a₁
=30−20
=10

a₃−a₂
=40−30
=10

a₄−a₃
=50−40
=10

As the differences are the same. So, the sequence is an arithmetic progression.
The graph represents a_n arithmetic progression given by 20,30,40,50,…

 

Page 125 Exercise 15 Answer

Given: Graph is

To find whether the representation is an arithmetic progression or not.
First, read the graph and write the sequence then check whether the difference between any two consecutive terms is the same or not.

By observing the graph, the ordered pairs are such that the first coordinate is the position of the term and the second coordinate is the value of the term.
So, the sequence will be a₁=21,​ a₂=18, a₃= 15,a₄=12

Checking whether the sequence is arithmetic progression or not.
Calculating the difference of all consecutive terms.
​a₂−a₁
=18−21
=−3

a₃−a₂
=15−18
=−3

a₄−a₃
=12−15
=−3
Hence, the sequence 21,18.15,12,…is an arithmetic progression.
The graph represents a_n arithmetic progression given by 21,18.15,12,…

 

Page 125 Exercise 16 Answer

Given sequence:−5.4,−6.6,−7.8,−9.0,…
To find the n^th term and a10

Substitute the values of the first term and common difference in the formula for n^th term and then substitute n=10i n the n^th term.

The given sequence is−5.4,−6.6,−7.8,−9.0,…
First-term a=−5.4
Common difference
​d=−6.6−(−5.4)​
=−6.6+5.4
=−1.2
​
The n^th term is ​a
=a+(n−1)d
=−5.4+(n−1)(−1.2)
=−5.4−1.2n+1.2
=−1.2n−4.2
​
Substitute n=10 in the n^th term, we get
​a₁₀=−1.2(10)−4.2
=−12−4.2
=−16.2
​
The equation for n^th  term of the sequence is−1.2n−4.2 and a10
=−16.2

 

Page 125 Exercise 17 Answer

 Given sequence: 43,38,33,28,…

To find the n^th term and a10
Substitute the values of the first term and common difference in the formula for n^th term and then substitute n=10 in the n^th term.

The given sequence is 43,38,33,28,…
First-term a=43
Common difference
​d=38−43
=−5

The n^th term is
​a_n=a+(n−1)d
=43+(n−1)(−5)
=43−5n+5
=−5n+48

​Substitute n=10 in the n^th term, we get
​a₁₀=−5(10)+48
=−50+48
=−2
​
The equation for n^th term of the sequence is−5n+48 and a₁₀
=−2

 

Page 125 Exercise 18 Answer

Given sequence: 6,10,14,18,…

To find the n^th term and a₁₀
Substitute the values of the first term and common difference in the formula for n^th term and then substitute n=10 in the n ^th term.

The given sequence is 6,10,14,18,…
First-term a=6
Common difference
​d=10−6
=4
​
The n^th term is
​an=a+(n−1)d
=6+(n−1)4
=6+4n−4
=4n+2
​
Substitute n=10 in the n^th term, we get
​a₁₀=4(10)+2
=42
​
The equation for the n^th term of the sequence is 4n+2 and a₁₀=42

 

Page 125 Exercise 19 Answer

Given sequence :−11,−9,−7,−5,…

To find the n^th term and a₁₀
Substitute the values of the first term and common difference in the formula for n^th term and then substitute n=10 in the n^th term.

The given sequence is−11,−9,−7,−5,…
First-term a=−11
Common difference
​d=−9−(−11)
=−9+11
=2

The n^th term is
​a_n=a+(n−1)d
=−11+(n−1)2
=−11+2n−2
=2n−13
​

Substitute n=10 in the n^th term, we get
​a₁₀=2(10)−13
=20−13
=7
​
The equation for n^th term of the sequence is 2n−13 and a₁₀=7

 

Page 125 Exercise 20 Answer

Given sequence:  34,37,40,43,…

To find the n^th term and a₁₀
Substitute the values of first term and common difference in the formula for n^th term and then substitute n=10 in the n^th term.

The given sequence is 34,37,40,43,…
First-term a=34
Common difference
​d=37−34
=3

The n^th term is
​a_n=a+(n−1)d
=34+(n−1)3
=34+3n−3
=3n+31
​
Substitute n=10 in the n^th term, we get
​a₁₀=3(10)+31
=61
​The equation for n^th term of the sequence is 3n+31 and a₁₀
=61

 

Page 125  Exercise 21  Answer

Given sequence is: \( \frac{9}{4}, \frac{7}{4}, \frac{5}{4}, \frac{3}{4}\)

To find the n^th term and a₁₀
Substitute the values of first term and common difference in the formula for n^th term and then substitute n=10 in the n^th term.

The given sequence is\( \frac{9}{4}, \frac{7}{4}, \frac{5}{4}, \frac{3}{4}\)

First-term a\(=\frac{9}{4}\)

Common difference
​d\(=\frac{7}{4}-\frac{9}{4}\)

\(=\frac{7-9}{4}\) \(=\frac{-2}{4}\) \(=\frac{-1}{2}\)

The n^th  term is ​a_n=a+(n−1)d

\(=\frac{9}{4}\)+(n−1)\(\left(\frac{-2}{4}\right)\)

\(\frac{9}{4}-\frac{2 n}{4}+\frac{2}{4}\) \(\frac{-n}{2}+\frac{11}{4}\)

Substitute n=10 in the n^th term, we get

​a₁₀ \(=\frac{-10}{2}+\frac{11}{4}\)

\(=\frac{-20+11}{4}\) \(=\frac{-9}{4}\)

​The equation for n^th  term of the sequence is \(\frac{-n}{2}+\frac{11}{4}\) and a₁₀=\(=\frac{-9}{4}\)