Pre-Calculus 11 Student Edition Chapter 1 Sequences and Series
Page 7 Problem 1 Answer
Follow the process of obtaining the staircase numbers mentioned above for each blank.
Complete the given box by the obtained values in the appropriate places.
For the third staircase number, add the number of cubes present in the third and fourth column.
Therefore, the third staircase number is 3+4=7.
Repeat the same process upto the tenth staircase number.
The fourth staircase number is4+5=9
The fifth staircase number is5+6=11
The sixth staircase number is6+7=13
The seventh staircase number is7+8=15
The eighth staircase number is8+9=17
The ninth staircase number is9+10=19
The tenth staircase number is10+11=21
Therefore the complete table is,
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The complete table is,
Page 7 Problem 2 Answer
Follow the process of obtaining the staircase numbers mentioned above for each blank.
Complete the given box by the obtained values in the appropriate places.
For the third staircase number, add the number of cubes present in the third, fourth and fifth column.
Therefore, the third staircase number is 3+4+5=12
Repeat the same process upto the tenth staircase number.
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The fourth staircase number is4+5+6=15
The fifth staircase number is5+6+7=18
The sixth staircase number is6+7+8=21
The seventh staircase number is7+8+9=24
The eighth staircase number is8+9+10=27
The ninth staircase number is9+10+11=30
The tenth staircase number is10+11+12=33
Therefore the complete table is,
The complete table is,
Page 8 Problem 3 Answer
The previously obtained numbers of cubes for a two step staircase are 3,5,7,9,11,…
If we observe we can see that each number is two more than the previous number.
Hence the pattern of the numbers is, ” Number of cubes in n+1th column = Number of cubes in nth column +2″
For a two step staircase, each number is two more than the previous number.
Hence the pattern of the numbers is, ” Number of cubes in n+1th column = Number of cubes in nth column +2″
Page 8 Problem 4 Answer
Following the mentioned rule, obtain the required data.
the number of cubes in the 11th term =11+12=23
And the number of cubes in the 12th term =12+13=25
the number of cubes in the 11th term =23
And the number of cubes in the 12th term =25
Page 8 Problem 5 Answer
Use the mentioned rule as the strategy to obtain the number of cubes required for staircases with three, four, five, or six steps.
Therefore, to obtain a particular three step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 3−1=2columns.
to obtain a particular four step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 4−1=3 columns.
to obtain a particular five step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 5−1=4 columns.
to obtain a particular six step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next6−1=5 columns.
to obtain a particular three step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 3−1=2 columns.
to obtain a particular four step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 4−1=3 columns.
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to obtain a particular five step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 5−1=4 columns.
to obtain a particular six step staircase number, obtain the sum of the number of cubes present in that particular staircase and in it’s next 6−1=5 columns.
Page 8 Problem 6 Answer
We know that, To obtain a particular n step staircase number, add the number of cubes present in that particular staircase and in it’s next n−1 columns.
and following the previous results we can conclude that, the difference between two consecutive numbers of the number of cubes is n in each case.
Which means the numbers follow a particular pattern of order, following which they can be considered a sequence with difference nbetween each consecutive terms.
Yes, the term of the number of cubes are considered as a sequence as, each consecutive terms has a difference of n.
Page 8 Problem 7 Answer
Following the previous results we can conclude that, the difference between two consecutive numbers of the number of cubes is n in each case.
Which means the numbers follow a particular pattern of order, following which they can be considered a sequence with difference n between each consecutive terms.
We observed a pattern of difference n between each consecutive terms. Specifically, each term is n more than it’s previous term.
Page 8 Problem 8 Answer
We know that, To obtain a particular n step staircase number, add the number of cubes present in that particular staircase and in it’s next n−1 columns and following the previous results we can conclude that, the difference between two consecutive numbers of the number of cubes is n in each case.
Therefore, for staircases of n steps (more than two steps), each term is generated by adding n more to the previous term.
For staircases of n steps (more than two steps), each term is generated by addingn
more to the previous term.
Page 8 Problem 9 Answer
To obtain a particular n step staircase number, add the number of cubes present in that particular staircase and in it’s next n−1 columns and each term can be generated by adding n to the previous terms.
Hence, the difference between two consecutive numbers of the number of cubes is n in each case.
Yes, the difference is same through out the whole sequence which is n for n staircase number.
Page 8 Problem 10 Answer
Follow the mentioned rule to obtain the 100th term of a two step staircase.
That is add the cubes present in the 100th and 101th column which results the addition of the numbers 100 and 101.
Total number of cubes
=100+101=201
To obtain the 100th term in a two-step staircase, we can find the sum of the 100th and 101th column which results the addition of the numbers 100 and 101, that is 201
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Page 8 Problem 11 Answer
Let, pn be the number of cubes in nth term.
If we convert the mentioned pattern in a formula, it becomes,
Then. pn=n+n+1=2n+1.
Substitute 100 for n in the above formula to obtain the 100th term.
So, for 100th term, the formula becomes, p100
=2⋅100+1
The formula by which the 100th term can be derived is p100
=2⋅100+1 where p100 is the 100th term.
Page 8 Problem 12 Answer
Let, the number of cubes in nth term be pn.
If we convert the mentioned pattern in a formula, it becomes,
pn=n+n+1=2n+1
The general formula to obtain the nth term is pn
=2n+1, where pn is the number of cubes in the nth term.
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Page 11 Problem 13 Answer
Given, the child was 70 cm tall at the age of 3 and we have to obtain the child’s height at any age starting from age 3, therefore, 70 can be considered as the first term. So, a=7.
The child increases 5 cm per year. Therefore the common difference is 5. So, d=5.
Now, the height at any age p, is the p−2 th term of the sequence, as the first term of the sequence occurs at age 3=1+2.
Substitute the above values in the formula mentioned to obtain the required formula.
At any age p, height of the child is tp−2=70+(p−2−1)5
tp−2=70+(p−3)5
tp−2=70+5p−15
tp−2=55+5p
At any age p, height of the child is, tp−2=55+5p.
Page 11 Problem 14 Answer
To obtain the height of the child at age 10, Substitute 10 for p in the mentioned formula and simplify it.
The child’s height at age 10 is ,
t10−2=55+5⋅10
t8=55+50=105
The child is expected to be of height 105 cm at age 10.
Page 14 Problem 15 Answer
Given that the 8th row contains 14 cans and the 12th row contains 10 cans. which implies, t8=14 and t12=10
Use the mentioned formula to substitute the values of t8 and t12 and simplify them to obtain the value of a in terms of d.
Next compare them to obtain the value of d.
Substitute the value of d in any of the values of d.
t8=14 implies, a+(8−1)⋅d=14
a+7d=14
a=14−7d
And t12=10
implies a+(12−1)d=10
a+11d=10
a=10−11d
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Comparing, 14−7d=10−11d
11d−7d=10−14
4d=−4
d=−1
Therefore substituting this value in the equation a=10−11d
we obtain,a=10−11⋅(−1)
a=10+11=21
Hence, tn=21+(n−1)(−1)
tn=21−n+1
tn=22−n
t1 is the first element which is already obtained as 21.
The final answers are,t1=21
d=−1
tn=22−n
Page 15 Problem 16 Answer
Given: Furnace technician charges $45, $46 per hour, for 10 hours
To find: The charge
For 10 hours the charge:
$460×10
=$460
Technician charge: $45
Total charge: $460+45=505
The total charge is $505.
Page 16 Problem 17 Answer
Given: t1=5 and d=3
We have to find the first four terms of the sequence.
Consider the first term of the sequence is t1=5 and d=3
Then, add the common difference to the previous term for the next term.
Therefore,
t2⇒5+3=8
t3⇒8+3=11
t4⇒11+3=14
Thus, the terms are 5,8,11,14.
Hence, the first four terms are 5,8,11,14.
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Page 16 Problem 18 Answer
Given:t1=−1 and d=−4
We have to find the first four terms of the sequence.
Consider the first term of the sequence is t1=−1 and d=−4
Then, add the common difference to the previous term for the next term.
Therefore,
t2⇒−1+(−4)=−5
t3⇒−5+(−4)=−9
t4⇒−9+(−4)=−13
Thus, the terms are−1,−5,−9,−13.
Hence, the first four terms are−1,−5,−9,−13.
Page 16 Problem 19 Answer
Given:t1=4 and d=1/5
We have to find the first four terms of the sequence.
Consider the first term of the sequence is t1=4 and d=1/5
Then, add the common difference to the previous term for the next term.
Therefore,
t2=4+1/5
t2=20+1/5
t2=21/5
t3=21/5+1/5
t3=22/5
t4=22/5+1/5
t4=23/5
Thus, the terms are 4,21/5,22/5,23/5.
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Hence, the first four terms are 4,21/5,22/5,23/5.
Page 16 Problem 20 Answer
Given : t1=1.25 and d=−0.25
We have to find the first four terms of the sequence.
Consider the first term of the sequence is t1=1.25 and d=−0.25
Then, add the common difference to the previous term for the next term.
Therefore,
t2⇒1.25+(−0.25)=1
t3⇒1+(−0.25)=0.75
t4⇒0.75+(−0.25)=0.5
Thus, the terms are1.25,1,0.75,0.5.
Hence, the first four terms are1.25,1,0.75,0.5.
Page 16 Problem 21 Answer
Given:tn=3n+8
We have to find t1
To find the indicated term, substitute the term number (n) in the given formula.
Consider tn=3n+8
Substitute n=1
Therefore,
t1=3(1)+8
t1=3+8
t1=11
Thus, t1=11.
Hence, the first term is t1=11.
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Page 16 Problem 22 Answer
Given:tn=3n+8
We have to find t7
To find the indicated term, substitute the term number (n) in the given formula.
Consider tn=3n+8
Substitute n=7
Therefore,
t7=3(7)+8
t7=21+8
t7=29
Thus, t7=29.
Hence, the seventh term is t7=29.
Page 16 Problem 23 Answer
Given:tn=3n+8
We have to find t14
To find the indicated term, substitute the term number (n) in the given formula.
Consider tn =3n+8
Substitute n=14
Therefore,
t14=3(14)+8
t14=42+8
t14=50
Thus, t14=50.
Hence, the 14th term is t14=50.
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Page 16 Problem 24 Answer
Given: The first term is 6 and the fourth term is 33.
We have to find the second and third term of the sequence.
Consider t1 =6 and t4 =33
Use formula tn=t1+(n−1)d
Substitute for t4.
t4=6+(4−1)d
33=6+3d
3d=33−6
3d=27
d=9
Now, Second term
t2=t1+d
t2=6+9
t2=15
t3=t2+9
t3=15+9
t3=24
Thus, the second term is 15 and the third term is 24.
Hence, the second term is 15 and the third term is 24.
Page 16 Problem 25 Answer
Given: The first term is 8 and the fourth term is 41.
We have to find the second and third term of the sequence.
Consider t1=8 and t4=41
Use formula tn
=t1+(n−1)d
Substitute for t4.
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t4=8+(4−1)d
41=8+3d
3d=41−8
3d=33
d=11
Now, t2=t1+d
t2⇒8+11=19
t3⇒19+11=30
Thus, the second term is 19 and the third term is 30.
Hence, the second term is 19 and the third term is 30.
Page 16 Problem 26 Answer
Given: The first term is 42 and the fourth term is 27
We have to find the second and third term of the sequence.
Consider t1=42 and t4=27
Use formula tn =t1+(n−1)d
Substitute for t4.
t4=t1+(4−1)d
27=42+3d
3d=27−42
3d=−15
d=−5
Now, t2⇒t1+d
t2⇒42+(−5)=37
t3⇒t2+d
t3⇒37+(−5)=32
Thus, the second term is 37 and the third term is 32.
Hence, the second term is 37 and the third term is 32.