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 Geometric Sequences

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisites:
(1.) Linear Systems
(2.) Exponents
(3.) 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 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 The Sunrise Preschool held its annual book fair for $3$ days.
The total profit for the $3$ days was $\$2,525$
The profit, in dollars, is shown for each of the $3$ days in the bar graph below.
Question 1 - GSA

The Sunrise Preschool plans to extend the book fair for an additional day.
To estimate the prfit for Day $4$, the preschool uses the geometric sequence that most closely approximates the daily profits for the $3$ days the book fair has been held.
Among the following, which is closest to the preschool's estimated profit for Day $4$?

$ A.\:\: \$0 \\[3ex] B.\:\: \$75 \\[3ex] C.\:\: \$140 \\[3ex] D.\:\: \$200 \\[3ex] E.\:\: \$275 \\[3ex] $

$ \underline{Profit} \\[3ex] Day\:1...clearly\:\:visible = \$1500 \\[3ex] \rightarrow first\:\:term = a = 1500 \\[3ex] Day\:2...not\:\:clear...skip\:\:for\:\:now \\[3ex] Day\:3...a\:\:little\:\:over\:\:\$300...assume\:\:\$310 \\[3ex] \rightarrow third\:\:term = 310 \\[3ex] first\:\:term + second\:\:term + third\:\:term = 2525...total\:\:profit \\[3ex] 1500 + second\:\:term + 310 = 2525 \\[3ex] 1810 + second\:\:term = 2525 \\[3ex] second\:\:term = 2525 - 1810 \\[3ex] second\:\:term = 715 \\[3ex] GS:\:\: 1500, 715, 310, fourth\:\:term \\[3ex] fourth\:\:term = Day\:4 \\[3ex] r = \dfrac{715}{1500} \\[5ex] fourth\:\:term = 310 * \dfrac{715}{1500} \\[5ex] fourth\:\:term = \dfrac{221650}{1500} \\[5ex] fourth\:\:term = 147.7666667 \\[3ex] Closest\:\:to\:\:the\:\:preschool's\:\:estimated\:\:profit\:\:for\:\:Day\:4 \approx \$140\:(based\:\:on\:\:the\:\:options) $
(2.) ACT The figure below shows the top view of the Santana family's house and yard.
The Santanas' rectangular house is $40$ feet wide and $30$ feet long, and their rectangular house is $75$ feet wide and $100$ feet long.
The Santanas have a rectangular garden in the back corner of their yard that is $30$ feet wide and $25$ feet long.
The garden currently contains $48$ flower bulbs: $10$ tulip bulbs, $18$ daffodil bulbs, and $20$ crocus bulbs.

Question 2 - GSA

Beginning next year, Mr. Santana will increase the number of bulbs in the garden each year so that the numbers form a geometric sequence.
In $3$ years, there will be $162$ bulbs in the garden.
By what factor will the number of bulbs be multiplied each year?

$ A.\:\: 1.125 \\[3ex] B.\:\: 1.5 \\[3ex] C.\:\: 3.375 \\[3ex] D.\:\: 4.85 \\[3ex] E.\:\: 38 \\[3ex] $

$ Based\:\:on\:\:the\:\:information: \\[3ex] 1st\:\:term = a = 48 \\[3ex] Increase\:\:each\:\:year\:\:until\:\:there\:\:are\:\:162\:\:bulbs\:\:in\:\:3\:\:years \\[3ex] GS:\:\:48,...,...,...,162 \\[3ex] 4th\:\:term = 162 \\[3ex] find\:\:r \\[3ex] GS_n = ar^{n - 1} \\[3ex] GS_4 = 48 * r^{4 - 1} \\[3ex] 162 = 48 * r^3 \\[3ex] 48 * r^3 = 162 \\[3ex] r^3 = \dfrac{162}{48} \\[5ex] r^3 = 3.375 \\[3ex] r = \sqrt[3]{3.375} \\[3ex] r = 1.5 \\[3ex] 2nd\:\:term = 48(1.5) = 72 \\[3ex] 3rd\:\:term = 72(1.5) = 108 \\[3ex] 4th\:\:term = 108(1.5) = 162...confirmed \\[3ex] $ The common ratio is $1.5$
This implies that each term needs to be multiplied by $1.5$ to get the subseqeunt term.
(3.)


(4.)


$ 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.)


(6.)


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.)


$ 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.)


$ 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.)


$ 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.)


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