Pre Calculus 11 Student Edition Chapter 1 Sequences and Series
Page 69 Problem 1 Answer
From the given incomplete arithmetic series, we see that the common difference is
d=9−3⇒d=6
So, the first term of the series will be a=3−6
⇒a=−3
The term after 9 will be 9+d=15
Similarly, the term after 15 will be 15+d=21
Hence, the sequence will be −3,3,9,15,21
Therefore, D.−3,15,21 will be the correct option.
We need to find the missing terms of the following arithmetic sequence:
The other options are not matching with the correct series.
Hence, D.−3,15,21 will be the correct option.
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Page 69 Problem 2 Answer
According to the given question, the number of cans in a row follows an arithmetic sequence with the first term a=1 and the common difference d=3.
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So, the number of cans in row n will be 1+(n−1)×3=3n−2
Hence, B.tn=3n−2 will be the correct option.
The grocery store display cans are arranged in the following order:
The first row contains 1 can and each successive row has 3 more cans than the previous row.
We need to find the correct option that would represent the number of cans in a row n.
Options A and D showed the total number of cans which is not correct.
Option C is not correct because the series do not satisfy the number of cans.
Hence, B.tn=3n−2 will be the correct option.
Page 69 Problem 3 Answer
To find sum of the geometric series 16807 − 2401 + 343 −…
t1 = 16807
n = 5
r = −2401/ 16807= − 1/7
Hence,
Sn =S5 = t1
( (r5 − 1 )/r − 1
S5 = 16807 ( (− 1/7 )5 − 1 )/(−1/7 − 1 )
S5 = 16807 (−1/16807 − 1 )/−1−7/7
S5 = 16807 (− 1 − 16807 )/16807/−8 /7
S5 = −16808 × 7/− 8
S5 = 2101 × 7
S5 = 14707
We have to find the sum of first five terms of geometric series 16807 − 2401 + 343
For option A. 19607 the sum we calculate does not match this option.
For option C. 16807.29 the sum we calculate does not match this option.
For option D. 14706.25 the sum we calculate does not match this option.
Hence the sum of first five terms of geometric series is 14707 which is
Option B 14707
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Page 69 Problem 4 Answer
Given sequence is arithmetic sequence whose first three terms are a,b and c.
We have to find third term i.e. c in terms of a and b.
For that we use formula for general term .
Given sequence is arithmetic sequence whose first three terms are a,b and c.
We have to find third term i.e. c in terms of a and b.
For that we use formula for general term .
For option A. a + b, it does not match answer calculated by general term formula.
For option C. a + ( n − 1 ) b, it does not match answer calculated by general term formula.
For option D. 2 a + b, it does not match answer calculated by general term formula.
Hence the third term c in terms of a and b is 2b −a which is Option B 2b − a
Page 69 Problem 5 Answer
We have to find the third term of the series and 20th term is given as 524288 and 14th term is 8192.
We apply the general term formula for 20th and 14th term and find the values of r and t1 and then we get the value of third term.
Given that 20th term i.e. t20 = 524288
14th term i.e. t14 = 8192
We use the formula for general term
For 20th term, n = 20
tn =t1 /rn−1
524288 =t1/r20−1
524288 =t1/r19
solve for t1
we get t1 = 524288
r19…………..(1)
For 14th term, n = 14
tn =t1 /rn−1
8192 = t1 /r14−1
8192 =t1/r13 ……………(2)
put the value of t1
from ( 1 ) in ( 2 )
8192 = 524288
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r19 ×r13
8192 = 524288/r6
r6 = 524288/8192
r6 = 64
r = 6/√64
r = 2
Substitute value of r in equation (1)
t1 = 524288/219
t 1 =524288/524288
t1 = 1
Hence for finding third term we put n= 3, t1 =1 and r =2 in general term we get,
t3 =t1
r3−1
t3 = 1 ×22
t3 = 4
We have to find third term of the geometric sequence whose 20th term is 524288 and 14th term is 8192.
We find 3rd term by applying general term formula.
For option B. 8 only , it does not match the answer calculated by general term formula.
For option C. + 4 or − 4, it does not match the answer calculated by general term formula.
For option D. + 8 or − 8, it does not match the answer calculated by general term formula.
Hence 3rd term of the sequence is 4 which is Option A. 4 only
Page 69 Problem 6 Answer
We have given the radius of the largest bowl i.e. 30 cm and for each successive bowl radius decreases by 90% and we have to find the radius of tenth bowl.
So we use formula for general term for n = 10 and r =0.90 and find the value of t10
i.e. radius of tenth bowl.
To find radius of tenth bowl, we use formula for general term.
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We have n = 10 , r =0.90,t1 = 30
tn =t1/rn−1
∴ t10
=30 × (0.90)10−1
t10 = 30 × (0.90)9
t10 = 30 × 0.3874
t10 = 11.62
Radius of tenth bowl is 11.62
Page 69 Problem 7 Answer
We have to compare the graph of arithmetic and geometric sequence.
Graphs are given to us and we have to find difference between two graphs.
By seeing above graphs
In the graph of arithmetic sequence all points are equally spaced while in geometric progression distance between two points is not same.
The difference between arithmetic and geometric sequence is that an arithmetic sequence has the difference between its two consecutive terms remains constant while a geometric sequence has the ratio between its two consecutive terms remains constant.
In the graph of arithmetic sequence all points are equally spaced while in geometric progression distance between two points is not same.
The difference between arithmetic and geometric sequence is that an arithmetic sequence has the difference between its two consecutive terms remains constant while a geometric sequence has the ratio between its two consecutive terms remains constant.
Page 70 Problem 8 Answer
Given: An arithmetic sequence,3, A,27.
A geometric sequence,3,B,27.
And, B>0.
To find: the values of A and B.
Since3, A,27 is an arithmetic sequence.
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Then,27−A=A−3
27+3=A+A
30=2A
2A=30
A=30/2
A=15
And,
When3,B,27 is a geometric sequence.
Then,27/B=B/3
By cross multiplication.
B2=27×3
B2=81
B=√81
B=±9
But,B>0
So,B=9
A=15,B=9
Page 70 Problem 9 Answer
Given that she walked 1700 km in 6 years and the percentage increase in the number of kilometers walked every week was 2%.
To find how many kilometeres does she walked in the first week.
Formula for sum of finite series is Sn=a1(rn−1)/r−1
Now, there are 52 weeks in a year this means there are312
weeks in 6 years.
Therefore, a1=?;r=100%+2%=1.02;n=312
17000=a1(1.02312−1)/1.02−1
17000⋅0.02=a1(1.02312−1)/340
(1.02312−1)=a1⋅(1.02312−1)/(1.02312−1)
a1=0.7065=0.71
Therefore, she walked around 0.71 km in the first week.
Page 70 Problem 10 Answer
Given: An arithmetic sequence,5,_, _, _,_,160.
To find: the unknown terms of the sequence.
From the sequence, it is obtained that.
n=6
a6=160
a1=5
Then,
an=a1+(n−1)d
a6=5+(6−1)d
160=5+5d
5d=160−5
5d=155
d=155/5
d=31
Hence,
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a2=a1+(2−1)31=5+(1×31)=5+31=36
a3=a1+(3−1)31=5+(2×31)=5+62=67
a4=a1+(4−1)31=5+(3×31)=5+93=98
a5=a1+(5−1)31=5+(4×31)=5+124=129
The complete sequence is 5,36,67,98,129,160.
Page 70 Problem 11 Answer
Given: An arithmetic sequence,5,_,_,_,_,160.
To find: the general term of the arithmetic sequence.
Note: From the solution of10(a),
the difference of the given sequence,d=31.
From the given sequence it is obtained that.
a1=5
Also,d=31
Then, the general terms of the given sequence are.
an=a1+(n−1)d=5+(n−1)31
The general terms of the given sequence are,an=5+(n−1)31.
Page 70 Problem 12 Answer
Given: a geometric sequence,5,_,_,_,_,160.
To find: the unknown terms of the sequence.
From the sequence, it is obtained that.
n=6
a6=160
a1=5
Then, an=a1.rn−1
a6=a1.r6−1
160=5×(r)5
r5=160/5
r5=32
r5=25
r=2
Hence,
a2=a1.(2)2−1
=5×(2)1
=5×2=10
a3=a1.(2)3−1
=5×(2)2
=5×4=20
a4=a1.(2)4−1
=5×(2)3
=5×8=40
a5=a1.(2)5−1
=5×(2)4
=5×16=80
The complete sequence is5,10,20,40,80,160.
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Page 70 Problem 13 Answer
the Given: First and last term of the geometric sequence is 5 and 160 respectively.
The first term of the geometric sequence is 5, so the value of a is 5.
From part d, the value of the common ratio, r is 2.
Determine the general term of the given geometric sequence.
an=5⋅2n−1
The general term of the geometric sequence is an=5⋅2n−1.
Page 70 Problem 14 Answer
Since the exposure time has a constant increment, the given problem can be solved by Arithmetic Progression.
The exposure time for the first day is 30s, therefore the first term of the series is a=30
The exposure increases by 30s each day. Therefore, the common difference of the terms is d=30.
Obtain first five terms by following the mentioned formula.
The second term is 30+30=60
The third term is 60+30=90
The fourth term is 90+30=120
The fifth term is 120+30=150.
Therefore the first five term of the sequence are 30,60,90,120,150.
The first five terms of the sequence are 30,60,90,120,150.
Page 70 Problem 15 Answer
For the given problem, the exposure increases at a constant rate of 30s each day.
Since the increment is constant, the given problem follows the Arithmetic progression.
The given sequence is Arithmetic.
Page 70 Problem 16 Answer
To reach the goal of 30 minutes, that is of 30×60=1800 seconds means to obtain the term with the term 1800.
For this problem, the first term a is 30
The common difference d is 30 the nth term tn is 1800
Substitute these values in the mentioned formula and simplify to obtain the value of the number of days a patient take to reach the goal that is the value of n
1800=30+(n−1)30
1800=30+30n−30
30n=1800
n=60
Therefore, on the 60 th day, the goal will be reached.
60 days are required to reach the goal.
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Page 70 Problem 17 Answer
To find the total number of minutes of the Sun exposure we have to find the sum of the sequence converted in seconds.
The first element is a=30
The common difference is d=30
The number of days are n=60
Substitute these values in the mentioned formula to obtain the final sum.
Sn=60/2(2⋅30+(60−1)30)
Sn=30(60+59⋅30)
Sn=30⋅30(2+59)
Sn=900⋅61
Sn=54900
Therefore the patient spends total 54900 seconds to reach the goal.
54900 seconds = 54900/60
=915minutes
The total number of Sun exposure is 915 when a patient reaches the goal.
Page 71 Problem 18 Answer
Gold can be an interesting natural resource that can be researched on.
In 1850’s British Columbia had a discovery of Gold. Later in 1896 the Klondike district of now known Yukon territory had a discovery of Gold which created a world wide rush in the demand and the industry of Gold.
The sequence of times when gold mines were discovered had a gap of on and average of 40 years.
The first term of this sequence can be assumed as 1856.
As the difference of the first and the second discoveries were about 40 years, assuming the discoveries in an Arithmetic Progression, we can assume that in next ten years, there might not be a great inventory of the production of Gold.Production of Gold has a big effect on the community as well as on the business, specially on the jewellery industry.
Gold is used in a huge amount to make fine ornaments as well as in recent days in clothings also.
Gold can be an interesting natural resource that can be researched on. In 1850’s British Columbia had a discovery of Gold.
Later in 1896 the Klondike district of now known Yukon territory had a discovery of Gold which created a world wide rush in the demand and the industry of Gold.
The sequence of times when gold mines were discovered had a gap of on and average of 40 years.
The first term of this sequence can be assumed as 1856.
As the difference of the first and the second discoveries were about 40 years, assuming the discoveries in an Arithmetic Progression, we can assume that in next ten years, there might not be a great inventory of the production of Gold.
Production of Gold has a big effect on the community as well as on the business, specially on the jewellery industry.
Gold is used in a huge amount to make fine ornaments as well as in recent days in clothings also.