2. Sums of Arithmetic Series
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Chapter 8
2.
Sums of Arithmetic Series
This lesson delves into arithmetic series and the sum formula, offering a step-by-step guide for calculating the sum of terms in such series. These series are sequences of numbers where the difference between each term is constant. The practical applications are diverse, from figuring out monthly loan payments to calculating the total number of gold bars in a stack. For instance, if one is considering a loan, understanding the sum formula can help you determine their monthly payments. Similarly, if you're faced with a task that involves counting items arranged in a specific pattern, like gold bars, the sum formula can be invaluable. The aim is to make these mathematical tools both accessible and useful for real-world situations.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Sums of Arithmetic Series
Slide of 11
In the case that a common difference occurs between the terms of a sequence, the terms represent a linear function. If the terms or the domain of the linear function are finite, then finding their sum makes sense in some cases. This lesson will introduce the way to find the finite sum of the terms of an arithmetic sequence.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Paying off a Loan
Diego's father has just taken a loan from the National Bank to start his own company. His first monthly payment will be $ 2400, after which the monthly payments will decrease by $100 each month — in other words, he will pay $2300 the second month, $ 2200 the third, and so on, until he pays a total of $ 24 500, including interest.
a How much will Diego's father pay off in total after five months?
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b How much will Diego's father pay off in total after ten months?
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c How long will it take Diego's father to pay off his total debt of $ 24 500?
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Discussion
Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence. If the sequence is finite and short enough, calculating the sum of its terms is quite straightforward.
| Example Arithmetic Sequence | 1, 3, 5, 7, 9 |
|---|---|
| Related Arithmetic Series | 1+ 3+ 5+ 7+ 9=25 |
However, if the sequence is still finite but longer, it can be tedious to add the terms by hand. In that case, the formula for an arithmetic sum can be used. If the arithmetic series is infinite, then the sum is said to diverge.
Example Divergent Arithmetic Series ∑_(n=1)^(∞) (3n+1) = 4+7+10+ ... ⇕ ∑_(n=1)^(∞) (3n+1) → ∞
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Pop Quiz
Identifying Arithmetic Series
The applet shows the first five terms of a sum. Identify whether the given sum is an arithmetic series or not.
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Discussion
Sum of an Arithmetic Series
For a finite arithmetic sequence with n terms and general formula a_n=a_1+(n−1)d, where a_1 is the first term and d the common difference, the sum of all terms S_n can be calculated using the following formula.
S_n=n(a_1+a_n)/2
Proof
Informal JustificationTo explain why the sum is determined this way, the sum of first nine natural numbers will be considered as an example for the arithmetic series.
1+2+3+⋯+9
Instead of adding all the terms in one go, pay close attention to the first term 1 and last term 9. The number 9 can be written as 4+5. Then, by the Associative Property of Addition, the sum 1+9 can be written as 5+5.
1+9
WriteSum
Write as a sum
1+4+5
AssociativePropAdd
Associative Property of Addition
(1+4)+5
AddTerms
Add terms
5+5
Therefore, the addition of the first and last terms is the same as 5+5. The sum of the second and second to last terms can be rewritten in the same manner. In fact, the sum of all opposite pairs can be rewritten as 5+5.
The sum of the 9 terms of the sequence is equivalent to adding nine 5s, which can be expressed as 9* 5. Note that here, 5 is the average of the first and last terms. This means that it can be written as half the sum of the first term a_1 and the last term a_n. Average of First and Last Terms a_1+a_n/2 To calculate the sum of the sequence's terms, which is the value of the related series S_n, this average is multiplied by the number of terms n. S_n=n* (a_1+a_n/2 ) [0.3em] ⇕ [0.3em] S_n= n(a_1+a_n)/2 It is worth mentioning that all arithmetic series with infinite terms diverge, meaning they tend towards infinity. Therefore, for an infinite arithmetic series, it only makes sense to use this formula to find a partial sum.
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Example
Calculating the Number of Tiles in the Restoration of the Floor
The board of the National Bank has decided to renew the old bank's floor. They decided to replace the old tiles with modern hexagonal tiles. The pavers start with a single tile and then form rings around it like a honeycomb. The diagram shows the progress of the work from the initial tile to the sixth ring.
a Write a rule to represent the number of tiles in the n^\text{th} ring. Note that the very first tile does not represent a ring.
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b Calculate the sum of the number of tiles after the first 10 rings are formed.
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c Calculate the sum of the number of tiles after the first 100 rings are formed.
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Hint
a Recall that the explicit rule for an arithmetic sequence is a_n=a_1+(n-1)d.
b Use the formula for the sum of an arithmetic series.
c Keep in mind that the initial tile does not represent a ring.
Solution
a Start by analyzing the number of tiles in each ring to identify the rule for the number of tiles in the n^\text{th} ring. Since the first tile does not represent a ring, it will not be considered for the sequence.
Since each ring has 6 more tiles than the previous one, the number of tiles in each ring represents an arithmetic sequence with common difference d= 6. Also, the first ring has 6 tiles. This means that a_1= 6. With this information, the explicit rule for the arithmetic sequence can be written.
a_n= a_1+(n-1) d
SubstituteII
a_1= 6, d= 6
a_n= 6+(n-1) 6
Distr
Distribute 6
a_n=6+6n-6
SubTerms
Subtract terms
a_n=6n
The number of tiles in the n^\text{th} ring can be represented with the rule a_n=6n.
b To calculate the total number of tiles after the 10^\text{th} ring is formed, the initial tile and the tiles in each of the ten rings will be added.
1+a_1+a_2+a_3+ ⋯ + a_9+a_(10) The number of tiles in each of the rings form an arithmetic series. Therefore, the sum of its first ten terms can be calculated by using the formula for the sum of an arithmetic series. S_n=n(a_1+a_n)/2 In this case, n= 10 will be substituted into the formula. S_(10)=10(a_1+a_(10))/2 Since it is already known that a_1= 6, the value of a_(10) will be calculated by using the general formula for this sequence, which was found in Part A.
There will be 60 tiles in the tenth ring. Now the value of S_(10) can be found.
This means that there are 330 tiles from the first to the tenth ring. Adding the 1 initial tile will be the last step to calculate the total number of tiles after the tenth ring is formed. 1+S_(10) ⇓ 1+330=331
c The total number of tiles in the first 100 rings will be calculated in a process very similar to that in Part B.
1+a_1+a_2+a_3+ ⋯ + a_(99)+a_(100) Once again, start by recalling the formula for the sum of an arithmetic series and substitute n= 100. S_n=n(a_1+a_n)/2 ⇓ S_(100)=100(a_1+a_(100))/2 Since there are a_1= 6 rings in the first round, the next step is to find the value of a_(100). To do so, use the general rule for this arithmetic sequence.
Now the formula for the sum of the arithmetic series can be used for n=100.
Finally, the initial tile will be added to this number. 1+S_(100) ⇓ 1+30 300 = 30 301 There will be 30 301 tiles making up the floor after the 100^\text{th} ring is formed.
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Example
Finding the Access Code
The National Bank has just created new security codes for its investors. When someone tries to log in, they have three tries to enter their code correctly. The third wrong attempt will block the account. Diego received an e-mail from the bank with information about this update.
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Hint
Use the formula for the sum of an arithmetic series.
Solution
Diego's access code is given as the sum of a series written in summation notation.
∑_(n=1)^(40) 2n+3
Here, the general rule of the series is a linear expression. This means that the series is arithmetic. With this in mind, the formula for the sum of an arithmetic series can be used.
S_n=n(a_1+a_n)/2
For Diego's case, n=40 will be substituted into the formula because the upper limit of the summation notation is 40. S_(40)=40(a_1+a_(40))/2
Next, the values of the first term a_1 and the last term a_(40) will be found one at a time. To do so, the general rule of the series will be used.
a_n=2n+3
Substitute
n= 1
a_1=2( 1)+3
IdPropMult
Identity Property of Multiplication
a_1=2+3
AddTerms
Add terms
a_1=5
The first term of the sequence is 5. The general rule will be used again to find the value of the last term a_(40).
The last term of the sequence is 83. Now the values of the first and last term can be used to calculate the sum of the series S_(40).
Diego's access code is 1760.
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Pop Quiz
Calculating the Sum of the Arithmetic Series
Calculate the sum of all the terms of the given finite arithmetic series written in summation notation. Use the formula for the sum of an arithmetic series.
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Example
Calculating the Number of Gold Bars
A vicious gang of thieves, famous for the masks they wear, is concocting a scheme to steal the National Bank's gold reserve.
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Before they make their move, they need to know how many bars of gold are kept in the reserve. According to the gang's mastermind, the bars of gold are stacked in the form of a triangle where each row has one more bar than the previous one.
a Write a rule to represent the number of bars in the n^\text{th} row of the triangular stack, supposing that there is one bar of gold in the top row.
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b Calculate the number of bars if the stack has 200 rows.
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c Calculate the number of bars in a triangular stack with 200 rows, supposing that there are 20 bars of gold in the top row.
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Hint
a The number of bars of gold in each row form an arithmetic sequence.
b Use the formula for the sum of an arithmetic series.
c Write a new general rule for the arithmetic sequence.
Solution
a Since each row has one more bar than the previous row, the number of bars in the rows form an arithmetic sequence. Knowing this, it is possible to write an explicit rule for the sequence.
a_n= a_1+(n-1) d Because the top row has one bar, the first term of this sequence is a_1= 1. Also, since each row has one more bar of gold than the previous one, the common difference is d= 1. With this information, the general rule for the sequence can be written.
a_n=a_1+(n-1)d
SubstituteII
a_1= 1, d= 1
a_n= 1+(n-1) 1
▼
Simplify right-hand side
IdPropMult
Identity Property of Multiplication
a_n=1+(n-1)
RemovePar
Remove parentheses
a_n=1+n-1
SubTerm
Subtract term
a_n=n
The number of bars of gold in each row is simply the number of the row.
b Since the number of bars in the stack increases by one, their sum represents an arithmetic series. This means that the sum can be calculated by using the formula for the sum of an arithmetic series.
S_n=n(a_1+a_n)/2 Now, n=200 will be substituted into the formula to find the sum of the number of bars in a stack with 200 rows. S_(200)=200(a_1+a_(200))/2 In this case, there is one bar of gold in the top row, so a_1= 1. The general rule of the sequence will be used to find the value of a_(200). a_n=n n = 200 a_(200)= 200 The last row of the stack has 200 bars of gold. Finally, the value of S_(200) can be calculated.
This means that there are 20 100 total bars of gold in the National Bank's reserve.
c In this case, the values of the first and last terms of the sequence are different than in Part B, so a new general rule must be defined. Since the top row of the triangular stack has 20 bars of gold, the first term is a_1= 20. The common difference is still d= 1. Now the explicit rule for the new arithmetic sequence can be written.
a_n=a_1+(n-1)d
SubstituteII
a_1= 20, d= 1
a_n= 20+(n-1)( 1)
▼
Simplify right-hand side
IdPropMult
Identity Property of Multiplication
a_n=20+(n-1)
RemovePar
Remove parentheses
a_n=20+n-1
SubTerm
Subtract term
a_n=n+19
Next, the value of the last term a_(200) will be calculated using the new general rule.
There are 219 bars in the 200th row. Finally, the total number of bars in the stack can be calculated by using the formula for the sum of an arithmetic series one more time. S_n=n(a_1+a_n)/2 ⇓ S_(200)=200(a_1+a_(200))/2 Substitute the values into the formula and simplify.
There are 23 900 bars of gold in this stack.
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Example
Calculating the Number of Police Officers
After the gang of thieves took over the bank, several police officers surrounded the building. The gang's mastermind needs to gather some information about the police officers in order to plan their escape.
a Write a rule for the number of police officers in the n^\text{th} row if there are 50 officers in the first row.
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b The gang's leader can see 15 rows of police. How many police officers are on the scene?
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Hint
a The numbers of police officers in each row represent an arithmetic sequence.
b Use the formula for the sum of an arithmetic series.
Solution
a The difference in the number of police officers between consecutive rows is constant, which means that the numbers represent an arithmetic sequence. Therefore, an explicit rule can be used to calculate how many police officers stand in each row.
a_n= a_1+(n-1) d It is given that each row has four more policemen than the previous row. Therefore, the common difference of the arithmetic sequence is d= 4. There are 50 officers in the first row. This means that the first term is a_1= 50. Now, substitute these values into the explicit rule.
a_n=a_1+(n-1)d
SubstituteII
a_1= 50, d= 4
a_n= 50+(n-1) 4
Distr
Distribute 4
a_n=50+4n-4
SubTerm
Subtract term
a_n=4n+46
The number of police officers in the n^\text{th} row is given by the general rule a_n=4n+46.
b The total number of police officers represents an arithmetic series. Therefore, the formula for the sum of an arithmetic series will be used.
S_n=n(a_1+a_n)/2 Since there are 15 rows, n=15 will be substituted into this formula. S_(15)=15(a_1+a_(15))/2 It is known that the first row has 50 officers, so a_1= 50. Therefore, only the number of police officers in the 15^\text{th} row needs to be found. To do so, the general formula for the arithmetic sequence will be used.
The 15th row has 106 policemen. Now, S_(15) will be calculated.
S_(15)=15(a_1+a_(15))/2
SubstituteII
a_1= 50, a_(15)= 106
S_(15)=15( 50+ 106)/2
AddTerms
Add terms
S_(15)=15(156)/2
Multiply
Multiply
S_(15)=2340/2
CalcQuot
Calculate quotient
S_(15)=1170
There are a total of 1170 police officers on the scene. It will be hard for the thieves to escape!
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Closure
Paying off a Loan
At the beginning of the lesson, Diego's father took out a loan from the National Bank to start his own company. He will start his repayment schedule by paying $ 2400 in the first month, after which the monthly amount will decrease by $100 each month. In the second month he will pay $ 2300, in the third month $ 2200, and so on.
a How much will Diego's father pay off in total after five months?
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b How much will Diego's father pay off in total after ten months?
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c How long will it take Diego's father to pay off his total debt of $ 24 500?
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Hint
a The amounts Diego's father will pay each month form an arithmetic sequence. The sum of these terms is an arithmetic series. Use the formula for the sum of an arithmetic series.
b As in Part A, use the formula for the sum of an arithmetic series.
c Solve the equation S_n=24 500 for n, where S_n represents the sum of the n terms of an arithmetic series.
Solution
a The first payment is $ 2400 and the monthly payments decrease by $ 100 each month. This means that the difference between payments is constant and the sequence has a common difference in the amount to be paid. Consequently, the monthly payments represent the terms of an arithmetic sequence.
a_n= a_1+(n-1) d Here, a_1 is the first term and d the common difference. In this case, since the payments decrease by $ 100 each month, the common difference is d= -100. Having the first payment a_1= $ 2400, the explicit rule can be written by substituting these values into the formula.
a_n=a_1+(n-1)d
SubstituteII
a_1= 2400, d= -100
a_n= 2400+(n-1)( -100)
Distr
Distribute -100
a_n=2400-100n+100
AddTerms
Add terms
a_n=- 100n+2500
The sum of the terms makes an arithmetic series. Therefore, the formula for the sum of an arithmetic series will be used to find the total amount that Diego's father pays off after 5 months. S_n=n(a_1+a_n)/2 ⇓ [0.5em] S_5=5(a_1+a_5)/2 To find S_5, the value of a_5 — the amount to be paid in the 5^\text{th} month — will be found by using the explicit rule.
a_n=- 100n+2500
Substitute
n= 5
a_5=- 100( 5)+2500
MultNegPos
(- a)b = - ab
a_5=- 500+2500
AddTerms
Add terms
a_5=2000
The amount to be paid in the fifth month is $ 2000. Now, substitute a_1=2400 and a_5=2000 into the formula to find S_5.
After 5 months, Diego's father will have paid a total of $ 11 000.
b This time the total amount of money paid will be calculated for 10 months.
S_(10)=10(a_1+a_(10))/2 First, n=10 will be substituted into the explicit rule of the corresponding arithmetic sequence to find a_(10).
This means that Diego's father will pay $ 1500 in the tenth month. Now S_(10) can be calculated by substituting a_1=2400 and a_(10)=1500 into the formula.
After ten months, Diego's father will pay off $ 19 500.
c To complete all his payments and pay off his loan completely, Diego's father has to pay a total of $ 24 500. The formula for the sum of an arithmetic series will be used to find how long it will take him to repay the loan completely.
S_n=24 500 ⇕ n(a_1+a_n)/2=24 500 To solve this equation, the explicit rule a_n=- 100n+2500 found in Part A and the first term of the sequence a_1=2400 will be substituted into the left hand side of the equation.
n(a_1+a_n)/2=24 500
SubstituteII
a_1= 2400, a_n= 2500-100n
n( 2400+ 2500-100n)/2=24 500
▼
Simplify
MultEqn
LHS * 2=RHS* 2
n(2400+2500-100n)=49 000
AddTerms
Add terms
n(4900-100n)=49 000
Distr
Distribute n
4900n-100n^2=49 000
AddEqn
LHS+100n^2=RHS+100n^2
4900n=49 000+100n^2
SubEqn
LHS-4900n=RHS-4900n
0=49 000+100n^2-4900n
RearrangeEqn
Rearrange equation
100n^2-4900n+49 000=0
FactorOut
Factor out 100
100(n^2-49+490)=0
DivEqn
.LHS /100.=.RHS /100.
n^2-49n+490=0
The equation is a quadratic equation. To solve it, the linear term will be rewritten as a difference.
n^2-49n+490=0
WriteDiff
Write as a difference
n^2-14n-35n+490=0
n(n-14)-35(n-14)=0
FactorOut
Factor out (n-14)
(n-14)(n-35)=0
Then, the Zero Product Property will be used.
(n-14)(n-35)=0
▼
Solve using the Zero Product Property
ZeroProdProp
Use the Zero Product Property
lcn-14=0 & (I) n-35=0 & (II)
AddEqn
(I): LHS+14=RHS+14
ln=14 n-35=0
AddEqn
(II): LHS+35=RHS+35
ln=14 n=35
There are two solutions to the equation. Notice that if the debt is completed in the 14^\text{th} month, which is before the 35^\text{th} month, there is no more debt to pay. Therefore, Diego's father will pay off his loan and interest for a total of $ 24 500 in 14 months.
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Sums of Arithmetic Series
Exercise 3.1
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{"type":"choice","form":{"alts":["Correct","Incorrect"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
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Let's analyze an arithmetic series with first term 1, a common difference of 1, and with only three terms. First Sum=1+2+3=6 Now let's create a new arithmetic series with the same number of terms and first term, but with a common difference of 5. This means that the second term is 1+5=6 and the third is 6+5=11. Second Sum=1+6+11=18 As we can see, the second sum is not five times as large as the first sum. Therefore, Emily's comment is incorrect.
We will check whether doubling each term in an arithmetic series doubles the sum of this series. Let S_n represent the arithmetic series whose terms will be doubled. S_n=a_1+a_2+...+a_n Now, let S_m be the series containing the double of each term of S_n. S_n=a_1+a_2+...+a_n ⇓ S_m=2a_1+2a_2+...+2a_n Notice that we can factor out 2 from S_m. S_m= 2a_1+ 2a_2+...+ 2a_n ⇕ S_m= 2(a_1+a_2+...+a_n) We can see that in the right-hand side of this equation, the multiplier is actually S_n. S_m=2( a_1+a_2+...+a_n) ⇓ S_m=2 S_n This means that if we multiply each term in an arithmetic series by 2, the sum of the new series is also multiplied by 2. Therefore, Emily's reasoning is correct.
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Sums of Arithmetic Series
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