Understanding Arithmetic Series: Your Guide to Sum Formula

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.

Arithmetic Sequence
Explicit Rule of Arithmetic Sequences
Series
Sigma Notation

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 $$ 24500,$ including interest.

External credits: Occupation vector created by freepik - www.freepik.com

a How much will Diego's father pay off in total after five months?

b How much will Diego's father pay off in total after ten months?

c How long will it take Diego's father to pay off his total debt of

$ 24500 ?

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 \sum \infty (3 n + 1) = 4 + 7 + 10 + \dots ⇕ n = 1 \sum \infty (3 n + 1) \to \infty

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.

Interactive applet showing different series

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.

$S_{n} = \frac{n ( a _{1} + a _{n} )}{2}$

Proof

Informal Justification

To 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 + \dots + 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

5 s,

which can be expressed as

9 \cdot 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 \frac{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 \cdot (\frac{a _{1} + a _{n}}{2}) ⇕ S_{n} = \frac{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.

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^{th}

ring. Note that the very first tile does not represent a ring.

b Calculate the sum of the number of tiles after the first

10

rings are formed.

c Calculate the sum of the number of tiles after the first

100

rings are formed.

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^{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 + 6 n - 6

SubTerms

Subtract terms

a_{n} = 6 n

The number of tiles in the

n^{th}

ring can be represented with the rule

a_{n} = 6 n .

b To calculate the total number of tiles after the

1 0^{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} + \dots + 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} = \frac{n ( a _{1} + a _{n} )}{2}

In this case,

n = 10

will be substituted into the formula.

S_{10} = \frac{1 0 ( 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.

a_{n} = 6 n

Substitute

$n = 10$

a_{10} = 6 (10)

Multiply

a_{10} = 60

There will be

60

tiles in the tenth ring. Now the value of

S_{10}

can be found.

S_{10} = \frac{1 0 ( a _{1} + a _{10} )}{2}

SubstituteII

$a_{1} = 6$ , $a_{10} = 60$

S_{10} = \frac{1 0 ( 6 + 6 0 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{10} = \frac{1 0 ( 6 6 )}{2}

Multiply

S_{10} = \frac{6 6 0}{2}

CalcQuot

Calculate quotient

S_{10} = 330

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} + \dots + a_{99} + a_{100}

Once again, start by recalling the formula for the sum of an arithmetic series and substitute

n = 100 .

S_{n} = \frac{n ( a _{1} + a _{n} )}{2} ⇓ S_{100} = \frac{1 0 0 ( 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.

a_{n} = 6 n

Substitute

$n = 100$

a_{100} = 6 (100)

Multiply

a_{100} = 600

Now the formula for the sum of the arithmetic series can be used for

n = 100 .

S_{100} = \frac{1 0 0 ( a _{1} + a _{100} )}{2}

SubstituteII

$a_{1} = 6$ , $a_{100} = 600$

S_{100} = \frac{1 0 0 ( 6 + 6 0 0 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{100} = \frac{1 0 0 ( 6 0 6 )}{2}

Multiply

S_{100} = \frac{6 0 6 0 0}{2}

CalcQuot

Calculate quotient

S_{100} = 30300

Finally, the initial tile will be added to this number.

1 + S_{100} ⇓ 1 + 30300 = 30301

There will be

30301

tiles making up the floor after the

10 0^{th}

ring is formed.

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.

Diego could really use some help. Help Diego to find his access code by calculating the sum!

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 \sum 40 2 n + 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} = \frac{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} = \frac{4 0 ( 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} = 2 n + 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} .

a_{n} = 2 n + 3

Substitute

$n = 40$

a_{40} = 2 (40) + 3

Multiply

a_{40} = 80 + 3

AddTerms

Add terms

a_{40} = 83

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} .

S_{40} = \frac{4 0 ( a _{1} + a _{40} )}{2}

SubstituteII

$a_{1} = 5$ , $a_{40} = 83$

S_{40} = \frac{4 0 ( 5 + 8 3 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{40} = \frac{4 0 ( 8 8 )}{2}

Multiply

S_{40} = \frac{3 5 2 0}{2}

CalcQuot

Calculate quotient

S_{40} = 1760

Diego's access code is

1760 .

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.

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.

External credits: user2122532

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^{th}

row of the triangular stack, supposing that there is one bar of gold in the top row.

b Calculate the number of bars if the stack has

200

rows.

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.

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} = \frac{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} = \frac{2 0 0 ( 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.

S_{200} = \frac{2 0 0 ( a _{1} + a _{200} )}{2}

SubstituteII

$a_{1} = 1$ , $a_{200} = 200$

S_{200} = \frac{2 0 0 ( 1 + 2 0 0 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{200} = \frac{2 0 0 ( 2 0 1 )}{2}

Multiply

S_{200} = \frac{4 0 2 0 0}{2}

CalcQuot

Calculate quotient

S_{200} = 20100

This means that there are

20100

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.

a_{n} = n + 19

Substitute

$n = 200$

a_{200} = 200 + 19

AddTerms

Add terms

a_{200} = 219

There are

219

bars in the

200 th

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} = \frac{n ( a _{1} + a _{n} )}{2} ⇓ S_{200} = \frac{2 0 0 ( a _{1} + a _{200} )}{2}

Substitute the values into the formula and simplify.

S_{200} = \frac{2 0 0 ( a _{1} + a _{200} )}{2}

SubstituteII

$a_{1} = 20$ , $a_{200} = 219$

S_{200} = \frac{2 0 0 ( 2 0 + 2 1 9 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{200} = \frac{2 0 0 ( 2 3 9 )}{2}

Multiply

S_{200} = \frac{4 7 8 0 0}{2}

CalcQuot

Calculate quotient

S_{200} = 23900

There are

23900

bars of gold in this stack.

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.

When the gang's leader peeked out, he noticed that the police officers are arranged in rows and that each row has four more officers than the previous row.

a Write a rule for the number of police officers in the

n^{th}

row if there are

50

officers in the first row.

b The gang's leader can see

15

rows of police. How many police officers are on the scene?

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 + 4 n - 4

SubTerm

Subtract term

a_{n} = 4 n + 46

The number of police officers in the

n^{th}

row is given by the general rule

a_{n} = 4 n + 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} = \frac{n ( a _{1} + a _{n} )}{2}

Since there are

15

rows,

n = 15

will be substituted into this formula.

S_{15} = \frac{1 5 ( 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

1 5^{th}

row needs to be found. To do so, the general formula for the arithmetic sequence will be used.

a_{n} = 4 n + 46

Substitute

$n = 15$

a_{15} = 4 (15) + 46

Multiply

a_{15} = 60 + 46

AddTerms

Add terms

a_{15} = 106

The

15 th

row has

106

policemen. Now,

S_{15}

will be calculated.

S_{15} = \frac{1 5 ( a _{1} + a _{15} )}{2}

SubstituteII

$a_{1} = 50$ , $a_{15} = 106$

S_{15} = \frac{1 5 ( 5 0 + 1 0 6 )}{2}

AddTerms

Add terms

S_{15} = \frac{1 5 ( 1 5 6 )}{2}

Multiply

S_{15} = \frac{2 3 4 0}{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!

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.

External credits: Occupation vector created by freepik - www.freepik.com

a How much will Diego's father pay off in total after five months?

b How much will Diego's father pay off in total after ten months?

c How long will it take Diego's father to pay off his total debt of

$ 24500 ?

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} = 24500

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 - 100 n + 100

AddTerms

Add terms

a_{n} = - 100 n + 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} = \frac{n ( a _{1} + a _{n} )}{2} ⇓ S_{5} = \frac{5 ( a _{1} + a _{5} )}{2}

To find

S_{5},

the value of

a_{5}

— the amount to be paid in the

5^{th}

month — will be found by using the explicit rule.

a_{n} = - 100 n + 2500

Substitute

$n = 5$

a_{5} = - 100 (5) + 2500

MultNegPos

$(- a) b = - a b$

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} .

S_{5} = \frac{5 ( a _{1} + a _{5} )}{2}

SubstituteII

$a_{1} = 2400$ , $a_{5} = 2000$

S_{5} = \frac{5 ( 2 4 0 0 + 2 0 0 0 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{5} = \frac{5 ( 4 4 0 0 )}{2}

Multiply

S_{5} = \frac{2 2 0 0 0}{2}

CalcQuot

Calculate quotient

S_{5} = 11000

After

5

months, Diego's father will have paid a total of

$ 11000 .

b This time the total amount of money paid will be calculated for

10

months.

S_{10} = \frac{1 0 ( a _{1} + a _{10} )}{2}

First,

n = 10

will be substituted into the explicit rule of the corresponding arithmetic sequence to find

a_{10} .

a_{n} = - 100 n + 2500

Substitute

$n = 10$

a_{10} = - 100 (10) + 2500

▼

Evaluate right-hand side

MultNegPos

$(- a) b = - a b$

a_{10} = - 1000 + 2500

AddTerms

Add terms

a_{10} = 1500

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.

S_{10} = \frac{1 0 ( a _{1} + a _{10} )}{2}

SubstituteII

$a_{1} = 2400$ , $a_{10} = 1500$

S_{10} = \frac{5 ( 2 4 0 0 + 1 5 0 0 )}{2}

▼

Evaluate right-hand side

AddTerms

Add terms

S_{10} = \frac{1 0 ( 3 9 0 0 )}{2}

Multiply

S_{10} = \frac{3 9 0 0 0}{2}

CalcQuot

Calculate quotient

S_{10} = 19500

After ten months, Diego's father will pay off

$ 19500 .

c To complete all his payments and pay off his loan completely, Diego's father has to pay a total of

$ 24500 .

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} = 24500 ⇕ \frac{n ( a _{1} + a _{n} )}{2} = 24500

To solve this equation, the explicit rule

a_{n} = - 100 n + 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.

\frac{n ( a _{1} + a _{n} )}{2} = 24500

SubstituteII

$a_{1} = 2400$ , $a_{n} = 2500 - 100 n$

\frac{n ( 2 4 0 0 + 2 5 0 0 - 1 0 0 n )}{2} = 24500

▼

Simplify

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

n (2400 + 2500 - 100 n) = 49000

AddTerms

Add terms

n (4900 - 100 n) = 49000

Distr

Distribute $n$

4900 n - 100 n^{2} = 49000

AddEqn

$LHS + 100 n^{2} = RHS + 100 n^{2}$

4900 n = 49000 + 100 n^{2}

SubEqn

$LHS - 4900 n = RHS - 4900 n$

0 = 49000 + 100 n^{2} - 4900 n

RearrangeEqn

Rearrange equation

100 n^{2} - 4900 n + 49000 = 0

FactorOut

Factor out $100$

100 (n^{2} - 49 + 490) = 0

DivEqn

$LHS / 100 = RHS / 100$

n^{2} - 49 n + 490 = 0

The equation is a quadratic equation. To solve it, the linear term will be rewritten as a difference.

n^{2} - 49 n + 490 = 0

WriteDiff

Write as a difference

n^{2} - 14 n - 35 n + 490 = 0

▼

Factor out

n & - 35

FactorOut

Factor out $n$

n (n - 14) - 35 n + 490 = 0

FactorOut

Factor out $- 35$

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

n - 14 = 0 n - 35 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 14 = RHS + 14$

n = 14 n - 35 = 0

AddEqn

$(II):$ $LHS + 35 = RHS + 35$

n = 14 n = 35

There are two solutions to the equation. Notice that if the debt is completed in the

1 4^{th}

month, which is before the

3 5^{th}

month, there is no more debt to pay. Therefore, Diego's father will pay off his loan and interest for a total of

$ 24500

14

months.

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Catch-Up and Review

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