If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Word Problems on Arithmetic Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) Quadratic Equations

Formulas: Formulas
Calculators: Calculators

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Use at least two methods whenever applicable.
Show all work

(1.) ACT A gardener is planting 7 rows of trees in a triangular plot.
The first row contains 1 tree.
Each successive row contains 2 more trees than the previous row.
How many trees will the gardener plant in the triangular plot?

$ F.\;\; 14 \\[3ex] G.\;\; 15 \\[3ex] H.\;\; 48 \\[3ex] J.\;\; 49 \\[3ex] K.\;\; 64 \\[3ex] $

$ 1st\;\;row = 1 \;\;tree \\[3ex] 2\;\;more\;\;trees\;\;than\;\;previous\;\;row \\[3ex] 2nd\;\;row = 1 + 2 = 3\;\;trees \\[3ex] 3rd\;\;row = 3 + 2 = 5\;\;trees \\[3ex] Arithmetic\;\;Sequence \implies \\[3ex] a = 1 \\[3ex] d = 2 \\[3ex] 7\;\;rows\;\;of\;\;trees \implies n = 7 \\[3ex] How\;\;many\;\;trees\;\;will\;\;be\;\;planted \implies Sum\;\;of\;\;all\;\;trees\;\;in\;\;all\;\;rows \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] SAS_7 = \dfrac{7}{2}[2(1) + 2(7 - 1)] \\[5ex] = \dfrac{7}{2}[2 + 2(6)] \\[5ex] = \dfrac{7}{2}[2 + 12] \\[5ex] = \dfrac{7}{2}(14) \\[5ex] = 7(7) \\[3ex] = 49 \\[3ex] $ The gardener will plant $49$ trees in triangular plot.
(2.) ACT Caden had exactly 45 plants to sell.
After Day 1 of his sale, he had exactly 42 plants left.
After Day 2, Caden had exactly 39 plants left.
After Day 3, he has exactly 36 plants left.
Assuming Caden will continue to sell plants at that daily rate, how many of these plants will he have left at the end of Day 6?

$ A.\;\; 33 \\[3ex] B.\;\; 27 \\[3ex] C.\;\; 24 \\[3ex] D.\;\; 6 \\[3ex] E.\;\; 3 \\[3ex] $

After:
Day $1$ ... $1st$ term = $45 - 3 = 42$ plants
Day $2$ ... $2nd$ term = $42 - 3 = 39$ plants
Day $3$ ... $3rd$ term = $39 - 3 = 36$ plants
Day $4$ ... $4th$ term = $36 - 3 = 33$ plants
As you can see, this is an arithmetic sequence
The question wants us to find the Day $6$ ... $6th$ term

Some people would just go ahead and complete it because the $6th$ term is not far away
Day $5$ ... $5th$ term = $33 - 3 = 30$ plants
Day $6$ ... $6th$ term = $30 - 3 = 27$ plants

Some people would prefer to use the formula:

$ a = 42 \\[3ex] d = 39 - 42 = -3 \\[3ex] n = 6 \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_{6} = 42 + -3(6 - 1) \\[3ex] AS_{6} = 42 + -3(5) \\[3ex] AS_{6} = 42 + -15 \\[3ex] AS_{6} = 42 - 15 \\[3ex] AS_{6} = 27 \\[3ex] $ Twenty seven plants will be remain after Day $6$
(3.)

(4.) ACT The manager of a grocery store asks Kamar to construct a display consisting of 10 rows of cans stacked on top of each other.
The manager wants the bottom row to have 25 cans and each succeeding row to have 1 less can than the row below it.
Part of 3 rows of the display is shown in the figure below.
How many cans will be in the top row of the display?

NUmber 4

$ A.\;\; 14 \\[3ex] B.\;\; 15 \\[3ex] C.\;\; 16 \\[3ex] D.\;\; 34 \\[3ex] E.\;\; 35 \\[3ex] $

Interpretation of the question:
The last term is the 25 cans. This is also known as the nth term
The common difference is 1 because each succeeding row has 1 less can than the row below it
The number of terms is 10 because the display consists of 10 rows of cans stacked on top of each other
We are asked to find the first term: How many cans will be in the top row of the display?

$ p = AS_n = 25 \\[3ex] n = 10 \\[3ex] d = 1 \\[3ex] a = ? \\[3ex] AS_n = a + d(n - 1) \\[3ex] 25 = a + 1(10 - 9) \\[3ex] 25 = a + 1(9) \\[3ex] 25 = a + 9 \\[3ex] a + 9 = 25 \\[3ex] a = 25 - 9 \\[3ex] a = 16 \\[3ex] $ 16 cans will be in the top row of the display
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(11.) ACT The 220 graduating seniors of Madison High School will sit in the center section of the school auditorium at the graduation ceremony. How many rows of seats will be needed to sit all of the graduating seniors if the first row has 10 seats and each succeeding row has 2 more seats than the previous row?

$ F.\;\; 10 \\[3ex] G.\;\; 11 \\[3ex] H.\;\; 12 \\[3ex] J.\;\; 15 \\[3ex] K.\;\; 30 \\[3ex] $

Teacher: This is an example of an Arithmetic Sequence. Student: Why is it an example of an Arithmetic Sequence?
Teacher: How many seats does the first row have?
Student: $10$ seats
Teacher: How many seats does the second row have?
Student: $12$ seats
Teacher: How did you get $12$ seats? Student: Because it says it from the question: each "succeeding" row has $2$ more seats than the previous row
So, it is $10 + 2 = 12$
Teacher: Correct!
How many seats does the third row have?
Student: $12 + 2 = 14$
That's right. It is an arithmetic sequence.
Teacher: What about the $220$ students? What variable is it?
Student: It is the sum of $n$ terms of the Arithmetic Sequence
Teacher: Why is that? Speaking like an American lol
Student: Because it is the total number of students that has to be seated
Teacher: Good answer
So, we have:

$ a = 10 \\[3ex] d = 2 \\[3ex] SAS_n = 220 \\[3ex] $ Are we missing anything
Student: $n$?
Teacher: And $n$ should be the:
Student: Number of rows...which is the number of terms ...
And that is what the question wants us to find out

We shall solve this in two ways.
Those two methods will use the $SAS_n$ formula
First Method: Formula - then - Factor
This method is recommended for the $ACT$ because the $ACT$ is a timed test

$ a = 10 \\[3ex] d = 2 \\[3ex] SAS_n = 220 \\[3ex] n = ? \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] 220 = \dfrac{n}{2}[2(10) + 2(n - 1)] \\[5ex] \dfrac{n}{2}[2(10) + 2(n - 1)] = 220 \\[5ex] LCD = 2 \\[3ex] Multiply\:\: both\:\: sides\:\: by\:\: 2 \\[3ex] n[2(10) + 2(n - 1)] = 2 * 220 \\[3ex] n(20 + 2n - 2) = 440 \\[3ex] n(2n + 18) = 440 \\[3ex] 2n^2 + 18n - 440 = 0 \\[3ex] Divide\:\: both\:\: sides\:\: by\:\: 2 \\[3ex] n^2 + 9n - 220 = 0 \\[3ex] (n + 20)(n - 11) = 0 \\[3ex] n = -20 \:\:OR\:\: n = 11 \\[3ex] $ The number of terms cannot be negative

$ \therefore n = 11 \\[3ex] $ There are $11$ rows of seats to seat the $220$ graduating seniors

Second Method: Formula - then - Formula (by Quadratic Formula)
Solved Examples - Literal Equations

$ a = 10 \\[3ex] d = 2 \\[3ex] SAS_n = 220 \\[3ex] n = ? \\[3ex] n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} \\[5ex] = \dfrac{-(2(10) - 2) \pm \sqrt{(2(10) - 2)^2 + 8 * 2 * 220}}{2 * 2} \\[5ex] = \dfrac{-(20 - 2) \pm \sqrt{(20 - 2)^2 + 3520}}{4} \\[5ex] = \dfrac{-18 \pm \sqrt{18^2 + 3520}}{4} \\[5ex] = \dfrac{-18 \pm \sqrt{324 + 3520}}{4} \\[5ex] = \dfrac{-18 \pm \sqrt{3844}}{4} \\[5ex] = \dfrac{-18 \pm 62}{4} \\[5ex] = \dfrac{-18 + 62}{4} \:\:OR\:\: \dfrac{-18 - 62}{4} \\[5ex] = \dfrac{44}{4} \:\:OR\:\: -\dfrac{80}{4} \\[5ex] = 11 \:\:OR\:\: -20 \\[3ex] $ The number of terms cannot be negative

$ \therefore n = 11 \\[3ex] $ There are 11 rows of seats to seat the 220 graduating seniors
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