Pre-Calculus 11 Student Edition Chapter 1 Sequences and Series
Page 33 Problem 1 Answer
When three coins are tossed, total outcomes is 8 ,expanded form is (2)(2)(2) and exponents is 23.
When four coins are tossed, total outcomes is 16, expanded form is (2)(2)(2)(2) and exponents is 24.
\(\begin{array}{c|c|c|c}\hline \begin{array}{c}
\text { Number of } \\
\text { Coins, } \boldsymbol{n}
\end{array} & \begin{array}{c}
\text { Number of } \\
\text { Outcomes, } \boldsymbol{n}_{\boldsymbol{n}}
\end{array} & \begin{array}{c}
\text { Expanded } \\
\text { Form }
\end{array} & \begin{array}{c}
\text { Using } \\
\text { Exponents }
\end{array} \\
\hline 1 & 2 & (2) & 2^1 \\
\hline 2 & 4 & (2)(2) & 2^2 \\
\hline 3 & & & \\
\hline 4 & & & \\
\hline \vdots & \vdots & \vdots & \\
\hline n & & & \\
\hline
\end{array}\)
Hence, the three , four coins are tossed the total outcomes will become 8 and 16.
Page 33 Problem 2 Answer
Given,
As the number of coins increases, a sequence is formed by the number of outcomes.
To find = What are the first four terms of this sequence?
If 1 coins is tossed, the total outcomes will become 2.
If 2 coins is tossed, the total outcomes will become 4.
If 3 coins is tossed, the total outcomes will become 8.
If 4 coins is tossed, the total outcomes will become 16.
So, the first four terms of this sequence is 2,4,8,16
Hence, the first four terms of this sequence is 2,4,8,16.
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Page 33 Problem 3 Answer
To find = Describe how the terms of the sequence are related. Is this relationship different from an arithmetic sequence? Explain.
An arithmetic sequence is a sequence where the difference d between successive terms is constant. …
An arithmetic series is the sum of the terms of an arithmetic sequence.
The nth partial sum of an arithmetic sequence can be calculated using the first and last terms.
An arithmetic sequence can be defined by an explicit formula in which an=a+(n−1)d, where d is the common difference between consecutive terms.
So, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. An arithmetic sequence is a list of numbers with a definite pattern.
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Page 33 Problem 4 Answer
Given, As the number of coins increases, a sequence is formed by the number of outcomes.
To find= Predict the next two terms of the sequence. Describe the method you used to make your prediction.
If 5 coins is tossed, the total outcomes will become 32.
If 6 coins is tossed, the total outcomes will become 64.
So, the next two terms of the sequence is 32 and 64.
The method is to first, find the common difference for the sequence.
Subtract the first term from the second term.
Subtract the second term from the third term.
To find the next value, add to the last given number.
Hence, the next two terms of the sequence is 32 and 64.
Page 33 Problem 5 Answer
Given: n coins are tossed.
To find: The method you could use to generate one term from the previous term.
For 2 coins, the outcomes are 22=4 and for 3 coins the total number of outcomes is 23=8.
Just need to multiply 2 with the total number of outcomes of the previous outcomes.
Need to multiply 2 to the previous term to get the term.
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Page 33 Problem 6 Answer
Given: n coins are tossed
To find: divide the second term by the preceding term.
N coins are tossed, and the total number of outcomes is 2n
n−1 coins are tossed, the total number of outcomes are 2n-1
Divide the terms:
2n/2n−1=2
Dividing the term with the previous term, we get 2.
Page 33 Problem 7 Answer
Given: n coins are tossed
To discuss: The prediction of the terms.
All we need to do is to multiply 2 by the previous term to get the term.
All we need to do is to multiply 2 by the previous term to get the term.
Page 35 Problem 8 Answer
Given: n coins are tossed
To find: divide the second term by the preceding term.
N coins are tossed, and the total number of outcomes is 2n
n−1 coins are tossed, the total number of outcomes are 2n−1
Divide the terms: 2n/2n-1=2
Dividing the term with the previous term, we get 2.
Page 33 Problem 9 Answer
Given: n coins are tossed
To discuss: About the prediction of the terms.
All we need to do is to multiply 2 to the previous term to get the trem.
All we need to do is to multiply 2 to the previous term to get the term.
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Page 35 Problem 10 Answer
Given, t1 = 10
t2 = 20
t3 = 40
To find = General term of the sequence.
Common ratio= 20/10=2
40/20=2
The common ratios is 2.
Use the geometric sequence
tn=t1/rn−1
tn =10(2)n−1
The general term of the sequence is tn
=10(2)n−1.
Page 35 Problem 11 Answer
Given,
First term= 42
Common ratio= 0.60
Number of terms= 9
Because you need to find the eighth term of the sequence.
tn =t1/rn−1
t9=42(0.60)9−1
t9=42(0.60)8
t9=0.705
Hence, after 8 reductions, the shortest possible length of a photograph is approximately 0.7cm.
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Page 37 Problem 12 Answer
Given, In a geometric sequence, the second term is 28 and the fifth term is 1792.
To find= t, r, and the first three terms of the sequence.
t2=28
t5=1792
t5=t2/r3
1792=28r3
1792/28=r3
r=4
t1=28/4= 7
tn =7(4)n−1
tn =28n−1
7,28,112….. is the first three terms of sequence.
Hence, the common ratio is 4 , the first number is 7 and the first three terms of sequence are 7, 28,112.
Page 39 Problem 13 Answer
Given, 1, 2, 4, 8, …
To find=.Determine if the sequence is geometric. If it is, state the common ratio and the general term in the form tn
=t1/rn−1
Common ratio: 2/1=2
4/2=2
8/4=2
Here, 2 is a common ratio.
Use the general term of a geometric sequence.
tn=t1/rn−1
tn=1(2)n−1
tn = 2n−1
Hence, the common ratio is 2 and general term of the sequence is tn =2n−1.
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Page 39 Problem 14 Answer
Given, 2,4,6,8
To find=.Determine if the sequence is geometric. If it is, state the common ratio and the general term in the form tn=t1/rn−1
The sequence is not geometric or arithmetic because there is no common difference or common ratio between each term.
Common ratios are = 4/2=2
6/4=3/2
8/6=4/3
Here, common ratios are different.
So this sequence is not geometric.
Hence, the sequence is not geometric or arithmetic.
Page 39 Problem 15 Answer
Given, 3, -9, 27, -81, …
To find= Determine if the sequence is geometric.
If it is, state the common ratio and the general term in the form
tn=t1/rn−1
Common ratio= −9/3=−3
= 27/−9=−3
=−81/27=−3
Here, common ratios is -3.
So, this sequence is geometric.
Use the general form
tn=t1/rn−1
tn=3(−3)n−1
tn =−9n−1
Hence, the common ratio is -3 and the general term is tn=−9n−1.
Page 39 Problem 16 Answer
Given, 1, 1, 2, 4, 8, …
To find = Determine if the sequence is geometric.
If it is, state the common ratio and the general term in the form
tn = t1/rn−1
Common ratio= 1/1=1
=2/1=2
4/2=2
8/4=2
Here, the common ratios are different.
The sequence is not geometric or arithmetic because there is no common difference or common ratio between each term.
Hence, the sequence is not a geometric sequence.
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Page 39 Problem 17 Answer
Given,
10, 15, 22.5, 33.75, …
To find = Determine if the sequence is geometric.
If it is, state the common ratio and the general
Look at the ratios between successive terms. If the ratios are all the same, the sequence is geometric. If not, it isn’t.
Common ratio = 15/10=1.5
22.5/15=1.5
33.75/22.5=1.5
Here, common ratios are 1.5.
The ratios are all the same, so this is a geometric sequence.
Use the general term,
tn =t1/rn−1
tn=10(1.5)n−1
tn =15n−1
Hence, the common ratios is 1.5 and the general term is tn=15n−1.
Page 39 Problem 18 Answer
Given, -1, -5, -25, -125, …
To find = Determine if the sequence is geometric. If it is, state the common ratio and the general form
Common ratio= −5/−1=5
−25/−5=5
−125/−25=5
Here, common ratio is 5.
So, this sequence is geometric.
Use the general form
tn =t1/rn−1
tn=−1(5)n−1
Hence, the common ratios is 5 and the general term is tn=−5n−1
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Page 39 Problem 19 Answer
Given; The table showing the first term.
\(\begin{array}{|c|c|c|c|}\hline \begin{array}{c}
\text { Geometric } \\
\text { Sequence }
\end{array} & \begin{array}{c}
\text { Common } \\
\text { Ratio }
\end{array} & \begin{array}{c}
\text { 6th } \\
\text { Term }
\end{array} & \begin{array}{c}
10 \text { th } \\
\text { Term }
\end{array} \\
\hline 6,18,54, \ldots & & & \\
\hline 1.28,0.64,0.32,- & & & \\
\hline \frac{1}{5}, \frac{3}{5}, \frac{9}{5},- & & & \\
\hline
\end{array}\)
We have to find the common ratio,6th term and 10 term.
Therefore the following table for the given geometric sequences.
\(\begin{array}{|c|c|c|c|}\hline \begin{array}{l}
\text { Ceometric } \\
\text { Sequence }
\end{array} & \begin{array}{l}
\text { Common } \\
\text { Ratio }
\end{array} & \begin{array}{l}
\text { 6th } \\
\text { Term }
\end{array} & \begin{array}{l}
\text { 10th } \\
\text { Term }
\end{array} \\
\hline 6,18,54, \ldots & 2 & 192 & 3,072 \\
\hline 1.28,0.64,0.32 \ldots & 0.5 & 0.04 & 0.0025 \\
\hline \begin{array}{l}
\frac{1}{5}, \frac{3}{5} \frac{9}{5},
\end{array} & 0.4 & 0.0020 & 5.2428 \\
\hline
\end{array}\)
Hence,Therefore the following table for the given geometric sequences.
Page 39 Problem 20 Answer
Given ; t1=2,r=3
To find the first four terms of each geometric sequence.
To find first four terms take n = 1,2,3 and 4 respectively.
⇒first term =2⋅31−1
=2⋅30
=2
⇒Second term =2⋅32−1
=2⋅31
=6
⇒Third term =2⋅33−1
=2⋅32
=2⋅9
=18
⇒ Fourth term =2⋅34−1
=2⋅27
=54
Hence , the first four terms of each geometric sequence are 2,6,18 and 54.
Page 39 Problem 21 Answer
Given ; t1=−3,r=−4
Determine the first four terms of each geometric sequence.
To find first four terms take n = 1,2,3 and 4 respectively.
⇒ First term =−3(−4)1−1
=−3(−4)0
=−3
⇒Second term =−3(−4)2-1
=−3(−4)
=12
⇒Third term =−3(−4)3−1
=−3(−4)2
=−3(16)
=−48
⇒Fourth term =−3(−4)4−1
=−3(−4)3
=−3(−64)
=192
Hence, the first four terms of each geometric sequence are −3,12,−48 and 192.
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Page 39 Problem 22 Answer
Given ; t1=4,r=−3
To Determine the first four terms of each geometric sequence.
To find the first four terms take n = 1,2,3 and 4 respectively.
⇒First term =4⋅(−3)1-1
=4⋅(−3)0
=4
⇒Second term =4⋅(−3)2-1
=4⋅(−3)
=−12
⇒Third term =4⋅(−3)3-1
=4⋅(−3)2
=36
⇒Fourth term =4⋅(−3)4-1
=4(−3)3
=4⋅(−27)
=−108
Hence, the first four terms of each geometric sequence are 4,−12,36 and −108.
Page 39 Problem 23 Answer
Given ; t1 =2, r=0.5
To find the first four terms of each geometric sequence.
To find the first four terms take n = 1,2,3 and 4 respectively.
⇒First term = 2⋅(0.5)1-1
=2⋅(0.5)0
=2
⇒Second term =2.(0.5)2-1
=2⋅(0.5)
=1
⇒Third term =2⋅(0.5)3-1
=2⋅(0.5)2
=2⋅0.25
=0.5
⇒Fourth term =2⋅(0.5)4-1
=2⋅(0.5)3
=2⋅(0.125)
=0.25
Hence, the first four terms of each geometric sequence are 2,1,0.5 and 0.25.
Page 39 Problem 24 Answer
Given; t1,t2,t3,t4,t5 in the geometric sequence.
⇒t1=8.1,t5
=240.1
To find the missing term: tn,t3,t4.
⇒ t4 = 240.1
⇒ 240.1 = 8.1(r)5-1
⇒ 240.1 = 8.1(r)4
⇒ 240.1/8.1= r4
⇒ 29.64 = r4
⇒ 2.3 = r
⇒ t2 = 8.1(2.3)2-1
= 8.1(2.3)
= 18.63
⇒ t3 = 8.1(2.3)3-1
= 8.1(2.3)2
= 8.1(5.29)
= 42.849
⇒ t4 = 8.1(2.3)4−1
= 8.1(2.3)3
= 8.1(12.16)
= 98.55
Hence, the missing terms are:t2
=18.63,t3
=42.849,t4
=98.55.
Page 39 Problem 25 Answer
Given ; r=2,t1=3
To Determine a formula for the nth term of each geometric sequence.
For the given sequence,t1=3, r=2
Use the general term of a geometric sequence.
⇒ tn =t1/rn−1
⇒ tn =(3)(2)n−1
=3(2)n−1
The general term of the sequence is tn=3(2)n−1.
Hence, formula for the nth term of each geometric sequence:tn
=3(2)n−1.
Page 39 Problem 26 Answer
Given ; 192, -48, 12, -3, …
Determine a formula for the nth term of each geometric sequence.
Common difference of the geometric progression is −48/192=−1/ 4
Use the general term of a geometric sequence.
⇒tn =t1/rn−1
⇒tn=192(−1/4)n−1
Hence, the general term of a geometric sequence is 192(−1/ 4)n−1.
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Page 39 Problem 27 Answer
Given :t3=5,t6=135
To Determine a formula for the nth term of each geometric sequence.
The third term of the sequence is 5 and sixth term is 135.
Since the sequence is geometric .
⇒ t4 =t3
r ⇒ t5
=t3/r⋅r
⇒ t6⋅r⋅r⋅r=t3
⇒ 135=5⋅r3
⇒ 135/5=r3
⇒ 27=r3
We have to find first term
⇒t3=t1/(3)3−1
⇒ 5=t1⋅9
⇒ t1 =5/9
The general formula for nth term is 5/9(3)n−1
Hence, the general formula for nth term is 5/9(3)n−1.
Page 39 Problem 28 Answer
For the given sequence, t1= 4, t131 = 16384
Determine a formula for the nth term of each geometric sequence.
⇒ t13 = t ⋅ r n−1
⇒ 16384 = 4 ⋅ r13−1
⇒16384/4= r12
⇒ 2.2 = r
Use general term for geometric sequence
⇒ tn = t1rn−1
⇒ tn = 4(2.2)n−1
Hence, the general formula for the nth term of each geometric sequence is :tn
=4(2.2)n−1.
Page 39 Problem 29 Answer
Given the following geometric sequences, determine the number of terms, n.
\(\begin{array}{c|c|c|c|}\hline \begin{array}{c}
\text { First } \\
\text { Term, } \\
\boldsymbol{t}_1
\end{array} & \begin{array}{c}
\text { Common } \\
\text { Ratio, } \boldsymbol{r}
\end{array} & \begin{array}{c}
\text { Term, } \\
\boldsymbol{t}_{\boldsymbol{n}}
\end{array} & \begin{array}{c}
\text { Number } \\
\text { of } \\
\text { Terms, } \boldsymbol{n}
\end{array} \\
\hline 5 & 3 & 135 & \\
\hline-2 & -3 & -1458 & \\
\hline \frac{1}{3} & \frac{1}{2} & \frac{1}{48} & \\
\hline 4 & 4 & 4096 & \\
\hline-\frac{1}{6} & 2 & -\frac{128}{3} & \\
\hline \frac{p^2}{2} & \frac{p}{2} & \frac{p^2}{256} & \\
\hline
\end{array}\)
Therefore the complete value of n;
\(\begin{array}{|c|c|c|c|}\hline \begin{array}{c}
\text { First } \\
\text { Term, } \\
\boldsymbol{t}_1
\end{array} & \begin{array}{c}
\text { Common } \\
\text { Ratio, } \boldsymbol{r}
\end{array} & \begin{array}{c}
\text { Term, } \\
\boldsymbol{t}_{\boldsymbol{n}}
\end{array} & \begin{array}{c}
\text { Number } \\
\text { of } \\
\text { Terms, } \boldsymbol{n}
\end{array} \\
\hline 5 & 3 & 135\\
\hline-2 & -3 & -1458\\
\hline \frac{1}{3} & \frac{1}{2} & \frac{1}{48}\\
\hline 4 & 4 & 4096\\
\hline-\frac{1}{6} & 2 & -\frac{128}{3}\\
\hline \frac{p^2}{2} & \frac{p}{2} & \frac{p^2}{256}\\
\hline
\end{array}\)