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It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
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The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

# Word Problems on Arithmetic Sequences

Prerequisites:
(1.) Linear Systems

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 any wrong answer.

For JAMB and CMAT 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 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 $6$
(2.)

$a = -6 \\[3ex] p = 71 \\[3ex] d = 2nd\:\: term - a \\[3ex] OR \\[3ex] d = 3rd\:\: term - 2nd\:\: term \\[3ex] = -2\dfrac{1}{2} - (-6) \\[5ex] = -\dfrac{5}{2} + 6 \\[5ex] = -\dfrac{5}{2} + \dfrac{12}{2} \\[5ex] \implies d = \dfrac{-5 + 12}{2} = \dfrac{7}{2} \\[5ex] OR \\[3ex] d = 1 - \left(-2\dfrac{1}{2}\right) \\[5ex] = 1 + \left(2\dfrac{1}{2}\right) \\[5ex] = 1 + \dfrac{5}{2} \\[5ex] = \dfrac{2}{2} + \dfrac{5}{2} \\[5ex] \implies d = \dfrac{2 + 5}{2} = \dfrac{7}{2} \\[5ex] n = \dfrac{p - a + d}{d} \\[5ex] n = (p - a + d) \div d \\[3ex] p - a + d = 71 - (-6) + \dfrac{7}{2} \\[5ex] = 71 + 6 + \dfrac{7}{2} \\[5ex] = 77 + \dfrac{7}{2} \\[5ex] = \dfrac{154}{2} + \dfrac{7}{2} \\[5ex] = \dfrac{154 + 7}{2} = \dfrac{161}{2} \\[5ex] \implies p - a + d = \dfrac{161}{2} \\[5ex] (p - a + d) \div d = \dfrac{161}{2} \div \dfrac{7}{2} \\[5ex] = \dfrac{161}{2} * \dfrac{2}{7} \\[5ex] = 23 \\[3ex] \implies n = 23$
(3.) 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?



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
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
(4.) JAMB The $3rd$ term of an arithmetic progression is $-9$ and the $7th$ term is $-29$.
Find the $10th$ term of the progression.

$AS_3 = a + 2d = -9 ...eqn.(1) \\[3ex] AS_7 = a + 6d = -29 ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \implies (a + 6d) - (a + 2d) = -29 - (-9) \\[3ex] a + 6d - a - 2d = -29 + 9 \\[3ex] 4d = - 20 \\[3ex] d = -\dfrac{20}{4} \\[5ex] d = -5 \\[3ex] Subst.\:\: d = -5 \:\:into\:\: eqn.(1) \\[3ex] a + 2(-5) = -9 \\[3ex] a - 10 = -9 \\[3ex] a = -9 + 10 \\[3ex] a = 1 \\[3ex] AS_{10} = a + 9d \\[3ex] AS_{10} = 1 + 9(-5) \\[3ex] AS_{10} = 1 - 45 \\[3ex] AS_{10} = -44$
(5.) JAMB The sixth term of an arithmetical progression is half of its twelfth term.
The first term is equal to
A. zero
B. half of the common difference
C. double the common difference
D. the common difference

$AS_6 = a + 5d \\[3ex] AS_{12} = a + 11d \\[3ex] AS_6 = \dfrac{1}{2} * AS_{12} \\[5ex] \implies a + 5d = \dfrac{1}{2}(a + 11d) \\[5ex] LCD = 2 \\[3ex] 2(a + 5d) = 2 * \dfrac{1}{2}(a + 11d) \\[5ex] 2a + 10d = a + 11d \\[3ex] 2a - a = 11d - 10d \\[3ex] a = d \\[3ex]$ The first term is equal to the common difference
(6.) ACT The second term of an arithmetic sequence is $12$, and the third term is $6$.
What is the first term?
(Note: In an arithmetic sequence, consecutive terms differ by the same amount.)

The ACT is a timed test - a question should typically take a minute to solve
We shall do it two ways
The first method is much faster. It is recommended for the ACT

$\underline{First\:\: Method - Faster} d = 3rd\:\:term - 2nd\:\: term \\[3ex] d = 6 - 12 = -6 \\[3ex] Also,\:\: d = 2nd\:\: term - 1st\:\: term \\[3ex] -6 = 12 - a \\[3ex] a = 12 + 6 \\[3ex] a = 18 \\[3ex] \underline{Second\:\: Method - Longer} \\[3ex] AS_n = a + d(n - 1) \\[3ex] AS_2 = a + d = 12 ...eqn.(1) \\[3ex] AS_3 = a + 2d = 6 ...eqn.(2) \\[3ex] 2 * eqn.(1) \implies 2(a + d) = 2(12) \\[3ex] 2 * eqn.(1) \implies 2a + 2d = 24...eqn.(3) \\[3ex] eqn.(3) - eqn.(2) \implies (2a + 2d) - (a + 2d) = 24 - 6 \\[3ex] 2a + 2d - a - 2d = 18 \\[3ex] a = 18 \\[3ex]$ The first term is $18$
(7.) WASSCE The second, fourth and sixth terms of an Arithmetic Progression ($AP$) are $x - 1$, $x + 1$, and $7$ respectively.
Find the
(i) common difference;
(ii) first term;
(iii) value of $x$

$AS_n = a + d(n - 1) \\[3ex] AS_2 = a + d = x - 1...eqn.(1) \\[3ex] AS_4 = a + 3d = x + 1...eqn.(2) \\[3ex] AS_6 = a + 5d = 7...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \implies (a + 3d) - (a + d) = (x + 1) - (x - 1) \\[3ex] a + 3d - a - d = x + 1 - x + 1 \\[3ex] 2d = 2 \\[3ex] d = \dfrac{2}{2} \\[5ex] (i)\:\: d = 1 \\[3ex] From\:\: eqn.(3) \\[3ex] a + 5d = 7 \\[3ex] a = 7 - 5d \\[3ex] Subst.\:\: d = 1 \\[3ex] a = 7 - 5(1) \\[3ex] a = 7 - 5 \\[3ex] (ii)\:\: a = 2 \\[3ex] From\:\: eqn.(1) \\[3ex] a + d = x - 1 \\[3ex] x - 1 = a + d \\[3ex] x = a + d + 1 \\[3ex] Subst.\:\: a = 2 \:\:and\:\: d = 1 \\[3ex] x = 2 + 1 + 1 \\[3ex] (iii)\:\: x = 4$
(8.) ACT The second term of an arithmetic sequence is $-11$, and the third term is $-38$.
What is the first term?
(Note: In an arithmetic sequence, consecutive terms differ by the same amount).

$AS_2 = a + d = -11 ...eqn.(1) \\[3ex] AS_3 = a + 2d = -38 ...eqn.(2) \\[3ex] From\:\: eqn.(1);\:\: d = -11 - a \\[3ex] Substitute\:\: (-11 - a)\:\:for\:\:d\:\:in\:\:eqn.(2) \\[3ex] a + 2(-11 - a) = -38 \\[3ex] a - 22 - 2a = -38 \\[3ex] -a = -38 + 22 \\[3ex] -a = -16 \\[3ex] a = \dfrac{-16}{-1} \\[5ex] a = 16$
(9.) Determine the first term and the common difference for an arithmetic sequence if the second term is $40 - 5n$ and the fourth term is $20m + n$

$AS_2 = a + d = 40 - 5n ...eqn.(1) \\[3ex] AS_4 = a + 3d = 20m + n ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \:\:gives \\[3ex] (a + 3d) - (a + d) = (20m + n) - (40 - 5n) \\[3ex] \rightarrow a + 3d - a - d = 20m + n - 40 + 5n \\[3ex] 2d = 20m + 6n - 40 \\[3ex] 2d = 2(10m + 3n - 20) \\[3ex] d = \dfrac{2(10m + 3n - 20)}{2} \\[5ex] d = 10m + 3n - 20 \\[3ex] eqn.(2) - 3 * eqn.(1) \:\:gives \\[3ex] (a + 3d) - [3(a + d)] = (20m + n) - [3(40 - 5n)] \\[3ex] \rightarrow a + 3d - (3a + 3d) = 20m + n - (120 - 15n) \\[3ex] a + 3d - 3a - 3d = 20m + n - 120 + 15n \\[3ex] -2a = 20m + 16n - 120 \\[3ex] -2a = 2(10m + 8n - 60) \\[3ex] a = \dfrac{2(10m + 8n - 60)}{-2} \\[3ex] a = -(10m + 8n - 60) \\[3ex] a = -10m - 8n + 60$
(10.)