Pre-Calculus 11 Student Edition Chapter 1 Sequences and Series
Page 46 Problem 1 Answer
Given:
To do: Complete the table.
In a fractal tree, every branch in each stage has two new branches.
In a fractal tree, every branch in each stage has two new branches.
Stage1 has one branch.Draw two line segments at the top of the segment in such a way that they are splitting away from each other.
Thus Stage 2 has two branches.
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Draw two line segments at the top of each segment of Stage 2 in such a way that they are splitting away from each other.
Since there are 2 branches in Stage 2, hence there will be 4 branches in Stage 3.
Similarly, Stage 4 has 8 branches and Stage 5 has 16 branches.
Now complete the table.
Hence the complete table is
Page 47 Problem 2 Answer
Definition: A geometric sequence is generated at each stage, after the first stage, is found by multiplying with the previous stage by a non-zero constant r, called the common ratio.
The general geometric sequence is t1,t1/r,t1/r2,t1/r3,………..
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The first term in the geometric sequence is t1
The common ratio is the ratio of successive terms of a geometric sequence.
The common ratio is calculated by dividing any two consecutive terms r=tn/tn−1
The general term of the geometric sequence is tn=t1/rn−1
Where tn is the general term r is the common ratio t1 is the first term n is number of terms
The first term in the geometric sequence is t1
The common ratio of the geometric sequence is r=tn/tn−1
The general term of the geometric sequence is tn=t1/rn−1
Page 47 Problem 3 Answer
To find: Determine whether the geometric sequence is used to find the total number of branches formed at the end of stage 5.
Given: The number of stages the branches formed is 5 (i.e)n=5.
The general term of the geometric sequences is tn=t1/rn−1
Yes, A geometric sequence can be generated.
By given t 5 be the total number of branches formed at the end of stage 5.
Let a be the initial number of branches (i.e) t1=a.
Let r be the common ratio.
Substituting in the geometric sequences we get,
t5=ar5−1
t5=ar4
Thus, The total number of branches formed at the end of stage 5 is t5=ar4.
Page 47 Problem 4 Answer
To find: Determine whether the geometric sequence is used to find the total number of branches formed at the end of stage 100.
Given: The number of stages the branches formed is 100 (i.e)n=100.
The general term of the geometric sequences is tn=t1/rn−1
Yes, A geometric sequence can be generated.
By given t100 be the total number of branches formed at the end of stage 100.
Let a be the initial number of branches (i.e) t1=a
Let r be the common ratio.
Substituting in the geometric sequences we get,
t100 =ar100−1
t100 =ar99
Thus, The total number of branches formed at the end of stage 100 is t100=ar99.
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Page 50 Problem 5 Answer
Sequence given in the question is,5+15+45+⋯
We will use the expression for the sum of 8 terms of the given geometric sequence.
Sn=a(rn−1)/r−1
For the given sequence,a=5
r=15/5=3
n=5
By substituting the values of a=5,r=3 and n=8 in the given expression,
Sn=a(rn−1)/r−1
S8=5(38−1)/3−1
=5(6561−1)/3−1
=5(6560)/2
=16400
Sum of 8 terms of the given geometric sequence (5+15+45+⋯) will be 16400.
Page 50 Problem 6 Answer
Expression for the sum of n terms of a geometric series → Sn=a(1−rn/)1−r
By substituting the values,
a=64,n=8,r=1/4
S8=64[1−(1/4)8]/1−1/4
=64[1−1/65536]/1−1/4
=64(65535/65536)/3/4
=64(65535)/65536×4/3
=5592320/65536
=21845/256
Sum of the 8 terms of the series with the first term 64,common ratio 1/4 will be 21845/256.
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Page 51 Problem 7 Answer
Series given in the question is, 1/64+1/16+1/4,…+1024
We will find the number of terms in the series with the expression,
Tn=arn−1 Where, Tn= nth term
a= First term
r= Common ratio
n= Number of term
Then we will use the expression for the sum of n terms of the series,
Sn=a(rn−1)/r−1
Where r>1
Given geometric series,
1/64+1/16+1/4+….+1024
Here, nth term of the series=1024
First terma=1/64
r=1/16
1/64 =64/16
=4
Substitute the values in the expression for nth term,
Tn=arn−1
1024=1/64(4n−1)/4n−1
=65536/4n−1
=48
n−1=8
n=9
Therefore, we have to find the sum of 9 terms of the given series.
Substitute the values in the expression for the sum of 9 terms of the given series,
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Sn=a(rn−1)/r−1
=1/64(49−1)/4−1
=(262144−1)/64×3
=262143/192
=87381/64
Sum of the given series (1/64+1/16+1/4+⋯+1024) will be 87381/64.
Page 51 Problem 8 Answer
Geometric series given in the question is,−2+4−8+….−8192
We will find the number of terms in the given series first.
Expression to get the number of terms in the series,
Tn=arn−1
Then we will find the sum of n terms by substituting the values of a,r and n in the expression,
Sn=a(1−rn)/1−r
where r<1
Given series in the question,−2+4−8+.…−8192
First terma=−2
r=4/−2
=−2 nth term of the series =−8192
Expression for the nth term of the given series,
Tn=arn−1
Substitute the values in the expression,−8192=(−2)(−2)n−1
4096=(−2)n−1/(−2)12
=(−2)n−1
12=n−1
n=13
We have to find the sum of 13 terms of the given series.
By using the expression,
Sn=a(1−rn)/1−r
Bu substituting the values a=(−2),r=−2,n=13,
S13=(−2)[1−(−2)13]/1−(−2)
=−2(1+8192)/1+2
=(−2)(8193)/3
=−5462
Sum of given series in the question in the question(−2+4−8+.…−8192) will be−5462.
Page 52 Problem 9 Answer
Number of participants in a tournament = 512
Therefore, in first round number of matches playedt1
=512/2
=256
In each round half the payers are eliminated, so the common ratio r=1/2
And the last term of the series will be final match, tn=1
By substituting these values in the expression, we can get the total matches played.
Expression for the sum of n terms of the geometric series,
Sn =rtn−t1/r−1
Here, tn=nth term of the series=1
t1=Matches played in first round=256
r=Common ratio=1/2
Substitute these values in the expression,
Sn=1/2(1)−256/1/2−1
=−511/2−1/2
=511
Total matches to be played with 512 participants will be 511.
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Page 53 Problem 10 Answer
The given series is 4+24+144+864+.………
We have to check whether the series is geometric or not.
Find the ratio of all terms to their preceding terms and check if the ratios are equal or not, if the ratio is equal then the given series will be a geometric series.
The given series is 4+24+144+864+.……….
The ratios are calculated as:
24/4=6/144
24=6/864
144=6
Since, the ratio of the term and their preceding term is constant throughout the series
Hence the given series is a geometric series.
The common ratio of the given series is 6, thus the given series is a geometric series.
Page 53 Problem 11 Answer
The given series is −40+20−10+5−.…
We have to check whether the series is geometric or not.
Find the ratio of all terms to their preceding terms and check if the ratios are equal or not, if the ratio is equal then the given series will be a geometric series.
The given series is −40+20−10+5−.……….
The ratios are calculated as:
20/(−40)=−1/2
(−10)/20=−1/2
5/(−10)=−1/2
Since, the ratio of the term and their preceding term is constant throughout the series.
Hence the given series is a geometric series.
The common ratio of the given series is −1/2, thus the given series is a geometric series.
Page 53 Problem 12 Answer
The given series is 3+9+18+54+.…..
We have to check whether the series is geometric or not.
Find the ratio of all terms to their preceding terms and check if the ratios are equal or not, if the ratio is equal then the given series will be a geometric series.
The given series is 3+9+18+54+.…
The ratios:9/3=3
18/9=2
54/18=3
Since the ratio of the term and their preceding term is not constant throughout the series.
Hence the given series is not a geometric series.
The ratio of the terms and their preceding terms are not constant in the given series, thus the given series is not a geometric series.
Page 53 Problem 13 Answer
The given series is 10+11+12.1+13.31+.…
We have to check whether the series is geometric or not.
Find the ratio of all terms to their preceding terms and check if the ratios are equal or not, if the ratio is equal then the given series will be a geometric series.
The given series is 10+11+12.1+13.31+⋯
Now the ratios are calculated as,
11/10=1.1
12.1/11=1.1
13.31/12.1=1.1
Since the ratio of the term and its preceding term is constant throughout the series.
Hence the given series is a geometric series.
The common ratio of the series is 1.1 so the given series is a geometric series.
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Page 53 Problem 14 Answer
We need to find sum of 10 terms if t1=12,r=2,n=10
Substitute these values in formula of sum of geometric series and solve.
S10 = 12(210 − 1)/2 − 1
=12(1024 − 1)/1
= 12(1023)
= 12276
The sum Sn for given geometric series is 12276
Page 53 Problem 15 Answer
We need to find sum Sn If t1=27,r=1/3,n=8
Substitute these values in formula of sum of geometric series and solve.
Sn =27[(1/3 )8 − 1]/1/3− 1
=27(1/6561 − 1)/−2/3
= −81(−6560)/2(6561)
=3280/81
The sum Sn for given geometric series is 3280/81.
Page 53 Problem 16 Answer
We are given
t1=1/256,r=−4,n=10
It is asked to find the Sn as exact values in fraction form.
The sum of the geometric series can be found by applying nth term sum formula.
We have
t1=1/256,r=−4,n=10
Apply the sum formula: Sn=t1(rn−1)/r−1
Plug values into the sum formula
S10=1/256((−4)10−1)/−4−1
S10=−209715/256
Hence, the sum in exact values is in fraction form S10=−209715/256
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Page 53 Problem 17 Answer
We are given t1=72,r=1/2,n=12
It is asked to find the Sn as exact values in fraction form.
The sum of the geometric series can be found by applying nth term sum formula.
We have t1=72,r=1/2,n=12
Apply the sum formula: Sn=t1(rn−1)/r−1
Plug values into the sum formula
S12=72((1/2)12−1)/1/2−1
S12=36855/256
Hence, the sum in exact values is in fraction form S12=36855/256
Page 53 Problem 18 Answer
We are given a geometric series 27+9+3+…..+1/243
It is asked to find the Sn in nearest hundredth
It can be found by finding the first term, common ratio, and the number of terms into the sum formula.
The series is 27+9+3+…..+1/243
Firstly, find the first term t1=27
Now, we can find the common ratio
The common ratio is the ratio of the second term and first term
r=9/27
r=1/3
Now, we can find the n
Apply general terms formula: tn
=t1(r)n−1
Plug these values into the formula
1/243=27(1/3)n−1
Solve for n(1/3)n−1=1/27×243(1/3)n−1
=1/33×35/(1/3)n−1
=1/38/(1/3)n−1
=(1/3)8
Compare exponent on both sides because bases are the same.
n−1=8
n=9
We have got
t1=27,r=1/3,n=9
Apply the sum formula: Sn=t1(rn−1)/r−1
Plug these values into the formula
S9=27((1/3)9−1)1/3−1
S9=9841/243
S9=40.50
Hence, the sum of geometric series is S9=40.50
Page 53 Problem 19 Answer
We are given a geometric series 1/3+2/9+4/27+……+128/6561
It is asked to find the Sn in nearest hundredth
It can be found by finding the first term, common ratio, and the number of terms into the sum formula.
The series is
1/3+2/9+4/27+……+128/6561
Firstly, find the first term
t1=1/3
Now, we can find the common ratio
The common ratio is the ratio of the second term and first term
r=2/9
1/3
r=2/3
Now, we can find the n
Apply general terms formula: tn
=t1(r){n−1}
Plug these values into the formula
128/6561=1/3(2/3)n−1
Solve for n(2/3)n−1
=3×128/6561(2/3)n−1
=128/2187(2/3)n−1
=27/37(2/3)n−1=(2/3)7
Compare exponent on both sides
n−1=7
n=8
We have got
t1=1/3,r=2/3,n=8
Apply the sum formula: Sn=t1(rn−1)/r−1
Plug these values into the formula
S8=1/3((2/3)8−1)/2/3−1
S8=6305/6561
S8=0.96
Hence, the sum is S8=0.96
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Page 53 Problem 20 Answer
We are given t1=5,tn=81920,r=4
It is asked to find the value of Sn
to the nearest hundred th The sum of the geometric series can be found by applying nth term sum formula.
We have
t1=5,tn=81920,r=4
Firstly, find the value of n
Apply general terms formula: tn=t1/(r){n−1}
Plug these values
81920=5(4){n−1}
5⋅4{n−1}=81920
5⋅4{n−1}
5=81920/5
4{n−1}=16384
4{n−1}=47Compare
exponent on both sides n−1=7
n=8
We have t1=5,tn
=81920,r=4,n=8
Apply the sum formula: Sn=t1(rn−1)/r−1
Plug these values into the formula
S8=5((4)8−1)/4−1
S8=109225
Hence, the sum of geometric series is S8=109225
Page 53 Problem 21 Answer
Given :-t1=3,tn=46875,r=−5
To find :- The sum of the series i.e.Sn.
To find the sum we have to first find number of terms in the series then put it in in formula of sum.
Put the given values in formula tn=t1/rn−1
46875=3(−5)n−1
⇒15625=(−5)n−1
⇒(−5)6
=(−5)n−1
having the same base so the power must be same
⇒n−1=6
n=7
So the number of term in the series are 7 put this value in second formula to obtain the sum.
Sn = t1(rn − 1)/r − 1
⇒ Sn = 3[(−5)7 − 1]/−5 − 1
⇒ Sn = 3[−78126]/−6
⇒ Sn = 39063
The sum of the given series is 39063
The sum of the given series is 39063.
Page 54 Problem 22 Answer
Given :-Sn=33,tn=48,r=−2
To find :- The first term of the given geometric series.
By using both formulas we will calculate the first term and number of terms in series.
First we will find out number of terms by dividing both the given formula.
Sn/an=a(rn−1)/r−1
a(rn−1)⇒sn
tn=(rn−1)
(r−1)(rn−1)⇒33
48=(−2)n−1−3(−2)n−1⇒33/16
=−(−2−1/(−2)n−1)⇒33/16
=2+1/(−2)n−1⇒33/16
−2=1/(−2)n−1⇒1/(−2)4
=1/(−2)n−1
⇒n−1=4
⇒n=5
Put this value of n in the formula we will get first term of the series.
tn=t1(rn−1)⇒48=t1/(−2)n−1
⇒48=t1(−2)5−1
⇒48=16t1
⇒t1=3
Hence we get the first term of the series is 3
The first term of the series is 3
Page 54 Problem 23 Answer
Given :-Sn=443,n=6,r=1/3
To find :- The first term of the series.
As the common ratio of the series is less than 1 and greater than -1 so we will use the given formula as in tip.
443 = t1(1 − (1/3 )6 )/1 − 1/3
⇒ 443 = t1(1 −1/729 )/2/3
⇒ 443 = t1x728 × 3/2 × 729
⇒ t1 =443 × 243/364
⇒ t1 = 295.73
⇒ t1 ≈ 296
The first term of the geometric series is 296
Page 54 Problem 24 Answer
Given :- The first term of the series is 4 and the common ratio of the series is 3 and also we have the sum of the series is 4372.
To find :- The number of the term of the series.
Put all the value in the formula 4372=4(3n−1)/3−1
⇒2186=3n−1
⇒3n
=2187
⇒3n
=37
We will compare the power since same base⇒n=7
The number of term in the geometric series are 7.
The number of term in the geometric series are 7.