Arithmetic Sequences and Series Word Problems with Step-by-Step Solutions

This is an arithmetic sequence case because the production increases by a fixed (constant) amount each month.
Given that \(a = 80\) and \(b = 10.\)
The total number of white-and-gray school uniforms produced during the first 6 months is
\(\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{6} & = \dfrac{6}{2}(2 \cdot 80 + (6-1) \cdot 10) \\ & = 3(160 + 50) \\ & = 3(210) = 630. \end{aligned}\)
Therefore, the total number of uniforms produced during 6 months is \(\boxed{630~\text{sets}}\)
(Answer D)

This is an arithmetic sequence case because production increases by a fixed (constant) amount each month.
Given that \(a = 8,000\) and \(b = 300.\)
The total number of goods produced during one semester (6 months) is
\(\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{6} & = \dfrac{6}{2}(2 \cdot 8,000 + (6-1) \cdot 300) \\ & =3(16,000 + 1,500) \\ & = 3(17,500) =52,500. \end{aligned}\)
Therefore, the total number of goods produced during one semester is \(\boxed{52,500~\text{units}}\)
(Answer C)

Since the difference between consecutive terms is constant, the case above is a contextual problem involving an arithmetic sequence.
Given \(\text{U}_1 = a = 50,000\) and \(b = 5,000.\)
We will find the value of \(\text{S}_{24}\) (2 years = 24 months).
$$\begin{aligned} \text{S}_n & = \dfrac{n}{2}(2a + (n-1)b) \\ \text{S}_{24} & = \dfrac{24}{2}(2 \times 50,000 + (24 -1) \times 5,000) \\ & = 12(100,000 + 115,000) \\ & = 12 \times 215,000 = 2,580,000 \end{aligned}$$Therefore, the total amount saved by the child over two years is Rp2,580,000.00.
(Answer B)

Given \(\text{U}_4 = 17\) and \(\text{S}_4 = 44\). Using the arithmetic series sum formula,
\(\boxed{\text{S}_n = \dfrac{n}{2}(a + \text{U}_n)}\)
we obtain
\(\begin{aligned} \text{S}_4 & = \dfrac{4}{2}(a + \text{U}_4) \\ 44 & = 2(a + 17) \\ \dfrac{44}{2} & = a + 17 \\ 22 & = a + 17 \\ a & = 5. \end{aligned}\)
Next, we determine the common difference \(b\).
\(\begin{aligned} \text{U}_4 = a + 3b & = 17 \\ 5 + 3b & = 17 \\ 3b & = 12 \\ b & = 4 \end{aligned}\)
Therefore, the production in the fifth month is
\(\boxed{\text{U}_5 = a + 4b = 5 + 4(4) = 21~\text{tons}}\)
(Answer D)

The amount of money saved by Asep each month forms an arithmetic sequence with first term \(a = 2,000\) and common difference \(b = 500\). In this case, we will find the value of \(\text{S}_{12}\) (1 year = 12 months).
$$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{12} & = \dfrac{12}{2}(2(2,000) + (12-1) \times 500) \\ & = 6(4,000 + 5,500) \\ & = 6(9,500) = 57,000 \end{aligned}$$Therefore, the total amount of money saved by Asep during the first year is Rp57,000.00.
(Answer D)

From the problem above, the number of seats in each row forms an arithmetic sequence with first term \(a = 20\), common difference \(b = 4\), and \(n=15.\)
Thus, we obtain
\(\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{15} & = \dfrac{15}{2}(2 \cdot 20 + (15-1) \cdot 4)\\ & = \dfrac{15}{2}(40 + 56) \\ & = \dfrac{15}{\cancel{2}} \cdot \cancelto{48}{96} \\ & = 15 \cdot 48 = 720. \end{aligned}\)
Therefore, the seating capacity of the theater is \(\boxed{720~\text{seats}}\)
(Answer C)

From the problem above, the annual production of Product A forms an arithmetic sequence with first term \(a = 1,960\), common difference \(b = -120\) (negative because production decreases), and \(n=16.\)
Thus, we obtain
$$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{16} & = \dfrac{\cancelto{8}{16}}{\cancel{2}}(2 \cdot 1,960 + (16 -1) \cdot (-120)) \\ & = 8(3,920 – 1,800) \\ & = 8 \cdot 2,120 = 16,960. \end{aligned}$$Therefore, the total production achieved up to the 16th year is \(\boxed{16,960~\text{units}}\)
(Answer C)

From the problem above, the monthly profit forms an arithmetic sequence with first term \(a = 46,000\), common difference \(b = 18,000\), and \(n=12.\)
Thus, we obtain
$$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{12} & = \dfrac{\cancelto{6}{12}}{\cancel{2}} (2 \cdot 46,000 + (12-1) \cdot 18,000) \\ & = 6(92,000 + 198,000) \\ & = 6(290,000) = 1,740,000. \end{aligned}$$Therefore, the trader’s total profit up to the 12th month is \(\boxed{\text{Rp}1,740,000.00}\)
(Answer A)

From the problem above, it is known that
\(\text{U}_1 = \dfrac{1}{5}\text{U}_{10} \Leftrightarrow 5\text{U}_1 = \text{U}_{10}\)
or it can be written as
\(5a = a + 9b \Leftrightarrow 20a = 45b.~~~~(1)\)
The sum of the ten central angles equals the total degrees in one complete circle, namely \(360^{\circ}\), so we write
$$\begin{aligned} \text{U}_1 + \text{U}_2 + \text{U}_3 + \cdots + \text{U}_{10} & = 360^{\circ} \\ a + (a + b) + (a + 2b) + \cdots + (a+9b) & = 360^{\circ} \\ 10a + (1+2+3+\cdots 9)b & = 360^{\circ} \\ 10a + 45b & = 360^{\circ}. && (\cdots 2) \end{aligned}$$Substitute equation \((1)\) into equation \((2)\).
\(\begin{aligned} 10a + 45b & = 360^{\circ} \\ 10a + 20a & = 360^{\circ} \\ 30a & = 360^{\circ} \\ \text{U}_1 & = a = 12^{\circ} \end{aligned}\)
The central angle of the largest pizza slice is
\(\text{U}_{10} = 5\text{U}_1 = 5(12^{\circ}) = 60^{\circ}.\)
The area of a sector with a central angle of \(60^{\circ}\) and radius \(\dfrac{20}{2} = 10~\text{cm}\) is
\(\begin{aligned} L & = \dfrac{60^{\circ}} {360^{\circ}} \pi r^2 \\ & = \dfrac{1}{6} \cdot 3.14 \cdot 100 \\ & = \dfrac{1}{6} \cdot 314 = 52\dfrac13. \end{aligned}\)
Therefore, the area of the largest pizza slice is \(\boxed{52\dfrac13~\text{cm}^2}\)
(Answer C)

The cattle populations form arithmetic sequences.
City A: Given \(a = 1,600\) and \(b = 25\), the cattle population in City A in month \(n\), counted from January 2019, is
\(\begin{aligned} A_n & = a + (n-1)b \\ & = 1,600 + (n -1) \cdot 25 \\ & = 1,575+ 25n. \end{aligned}\)
City B: Given \(a = 500\) and \(b = 10\), the cattle population in City B in month \(n\), counted from January 2019, is
\(\begin{aligned} B_n & = a + (n-1)b \\ & = 500 + (n-1) \cdot 10 \\ & = 490 + 10n. \end{aligned}\)
Since the cattle population in City A is three times that of City B, we obtain
\(\begin{aligned} A_n & = 3B_n \\ 1,575 + 25n & = 3(490 + 10n) \\ 1,575 + 25n & = 1,470 + 30n \\ 5n & = 105 \\ n & = 21. \end{aligned}\)
This means that 21 months after January 2019, the cattle population in City A will become three times the cattle population in City B. The cattle population in City A is \(A_{21} = 1,600 + (21-1) \cdot 25 = 2,100\) head.
(Answer D)

Given:
\(a = 20, n = 30\), and \(\text{S}_n = 22,350.\)
Asked: \(\text{U}_n\)
Using the arithmetic series sum formula, we obtain
\(\begin{aligned} \text{S}_n & = \dfrac{n} {2}(a + \text{U}_n) \\ 22,350 & = \dfrac{30}{2}(20 + \text{U}_n) \\ 22,350 & = 15(20 + \text{U}_n) \\ 20 + \text{U}_n & = \dfrac{22,350}{15} = 1,490 \\ \text{U}_n & = 1,490 -20 = 1,470. \end{aligned}\)
Therefore, the number of eggs he collected on the last day was \(\boxed{1,470~\text{eggs}}\)
(Answer D)

Based on the given information, the amounts of money received by the children form an arithmetic sequence with common difference \(b = 10,000\) and \(\text{S}_5 = 200,000.\)
Using the arithmetic series sum formula, we obtain
$$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_5 & = \dfrac{5}{2}(2a + (5-1)\cdot 10,000) \\ 200,000 & = \dfrac{5}{2}(2a + 40,000) \\ 200,000 \cdot \dfrac25 & = 2a + 40,000 \\ 80,000 & = 2a + 40,000 \\ 2a & = 40,000 \\ a & = 20,000. \end{aligned}$$The value of \(\text{U}_3\) is given by
\(\begin{aligned} \text{U}_3 & = a + 2b \\ & = 20,000+2(10,000) = 40,000. \end{aligned}\)
Therefore, the 3rd child receives Rp40,000.00.
(Answer C)

This is an arithmetic sequence case.
From the position of bottle number 10, the participant moves toward the box for a distance of \(9 \times 8 + 10 = 82~\text{m}.\)
Starting from the box position:
\(\text{U}_1\) is the distance from the box to bottle number 1.
\(\text{U}_2\) is the distance from the box to bottle number 2, and so on, so that
\(\text{U}_1 = 10,\) \(\text{U}_2 = 18,\) \(\text{U}_3 = 26,\) up to \(\text{U}_{10} = 10 + 8 \times 9 = 82.\)
Therefore, the total round-trip distance traveled by the participant (hence multiplied by two) is
\(\begin{aligned} 2 \text{S}_n & = 2 \times \dfrac{n}{2}(\text{U}_1 + \text{U}_n) \\ 2 \text{S}_{10} & = 10(10 + 82) \\ & = 10 \times 92 = 920. \end{aligned}\)
However, note that when the participant carries the last flag and rushes to bottle number 10, there is no need to return to the box because the game has already been completed. Therefore, the total distance traveled is $$\boxed{s = 82 + 920 – (8 \times 9 + 10) = 920~\text{m}}$$(Answer C)

Suppose the cheapest item has price \(\text{U}_1\).
Next, we obtain
\(\begin{aligned} \text{U}_1 + \text{U}_2 + \text{U}_3 + \text{U}_4 & = 50 \\ \text{U}_4 + \text{U}_5 + \text{U}_6 + \text{U}_7 & = 86. \end{aligned}\)
If expressed in the form \(\text{U}_n = a + (n-1)b\), we obtain
$$\begin{aligned} a + (a + b) + (a+2b) + (a+3b) & = 50 && (\cdots 1) \\ (a+3b) + (a+4b) + (a+5b) + (a+6b) & = 86 && (\cdots 2) \end{aligned}$$Simplify.
\(\begin{aligned} 4a + 6b & = 50 && (\cdots 1) \\ 4a + 18b & = 86 && (\cdots 2) \end{aligned}\)
Subtract the two equations above to obtain \(b = 3\), which implies \(a = 8.\)
Therefore, the prices of the seven goods are \(8, 11, 14, 17, 20, 23,\) and \(26.\)
If the customer buys items priced at \(14, 17, 20, 23,\) and \(26\) (total: \(100\)), then the payment is exact with no change.
Therefore, the minimum change received is \(\boxed{0}\)
(Answer A)

Consider the sequence \(2, 5, 8, 11, 14, \cdots\), which is an arithmetic sequence with first term \(a = 2\) and common difference \(b = 3.\) The formula for the \(n\)-th term is
$$\begin{aligned} \text{U}_n & = a + (n-1)b \\ & = 2+(n-1) \cdot 3 \\ & = 3n-1. \end{aligned}$$Since Wesley says the number \(41,\) we obtain
$$\begin{aligned} 41 & = 3n-1 \\ 42 & = 3n \\ n & = 14. \end{aligned}$$Therefore, Wesley is in the 14th position from the left. This means there are 13 children to Wesley’s left.
Now consider the sequence \(1, 5, 9, 13, 17, \cdots\), which is an arithmetic sequence with first term \(a = 1\) and common difference \(b = 4.\) The formula for the \(n\)-th term is
$$\begin{aligned} \text{U}_n & = a + (n-1)b \\ & = 1+(n-1) \cdot 4 \\ & = 4n-3. \end{aligned}$$Since Wesley says the number \(41,\) we obtain
$$\begin{aligned} 41 & = 4n-3 \\ 44 & = 4n \\ n & = 11. \end{aligned}$$Therefore, Wesley is in the 11th position from the right. This means there are 10 children to Wesley’s right.
Thus, the total number of children is \(\boxed{13 + 1 + 10 = 24}\)
(Answer D)

Since the fare increases by a constant amount, this is a contextual problem involving an arithmetic sequence.
Given \(\text{U}_1 = a = 6,000\) and \(b = 2,500.\) In this case, we are asked to find the value of \(\text{U}_{15},\) which represents the taxi fare for a 15 km trip.
$$\begin{aligned} \text{U}_{n} & = a+(n-1)b \\ \text{U}_{15} & = 6,000 + (15-1) \cdot 2,500 \\ & = 6,000 + 35,000 \\ & = 41,000. \end{aligned}$$Therefore, the taxi fare that the passenger must pay for a 15 km trip is Rp41,000.00.

Since the average growth rate is assumed to be constant, this is a contextual problem involving an arithmetic sequence.
The plant population in the first year is \(\text{U}_1 = a = 75,230.\) In the eighth year, the population becomes \(\text{U}_8 = a + 7b = 125,280,\) where \(b\) is the average growth rate. Substituting \(a = 75,230\) into \(\text{U}_8\) gives
$$\begin{aligned} 75,230 + 7b & = 125,280 \\ 7b & = 50,050 \\ b & = 7,150. \end{aligned}$$Therefore, the average population growth is \(\boxed{7,150}.\)

Given \(\text{U}_1 = a = 75\) and \(b = 125-75 = 50.\) In this case, we are asked to find \(\text{S}_{20},\) the sum of the first 20 terms of an arithmetic sequence with first term 75 and common difference 50.
$$\begin{aligned} \text{S}_{n} & = \dfrac{n}{2}(2a+(n-1)b) \\ \text{S}_{20} & = \dfrac{20}{2}(2(75) + (20-1) \cdot 50) \\ & = 10(150 + 950) \\ & = 11,000. \end{aligned}$$Therefore, the total number of oranges harvested during the first 20 days is \(\boxed{11,000}\) oranges.

Since the number of seats increases by a constant amount, this is a contextual problem involving an arithmetic series.
Given \(\text{U}_1 = a = 15\) and \(b = 20-15 = 5.\) In this case, we are asked to find \(\text{S}_{20},\) the sum of the first 20 terms of an arithmetic sequence with first term 15 and common difference 5.
$$\begin{aligned} \text{S}_{n} & = \dfrac{n}{2}(2a+(n-1)b) \\ \text{S}_{20} & = \dfrac{20}{2}(2(15) + (20-1) \cdot 5) \\ & = 10(30 + 95) \\ & = 1,250. \end{aligned}$$Therefore, the total number of seats available in the auditorium is \(\boxed{1,250}\) seats.

Since the decrease in distance is constant, this is a contextual problem involving an arithmetic series.
Given \(\text{U}_1 = a = 1,150\) and \(b = -75\) (negative because the distance decreases). Since 3 months equal 12 weeks, we are asked to find \(\text{S}_{12},\) the sum of the first 12 terms of an arithmetic sequence with first term 1,150 and common difference \(-75.\)
$$\begin{aligned} \text{S}_{n} & = \dfrac{n}{2}(2a+(n-1)b) \\ \text{S}_{12} & = \dfrac{12}{2}(2(1,150) + (12-1) \cdot (-75)) \\ & = 6(2,300-825) \\ & = 6 \cdot 1,475 \\ & = 8,850. \end{aligned}$$This means that the salesperson has traveled a total distance of 8,850 km during the 3-month period. Since each liter of gasoline allows the motorcycle to travel 30 km, he needs \(\dfrac{8,850}{30} = 295\) liters. Since one liter of gasoline costs Rp10,000.00, the amount of money he must spend on gasoline is \(295 \times\) Rp10,000.00, which is Rp2,950,000.00.

Since the increase in roof tile production each day is constant, this is a contextual problem involving an arithmetic series. Note that 1 dozen is equal to 12 pieces.
Given \(\text{U}_1 = a = 5 \cdot 12 = 60\) and \(b = 1 \cdot 12,\) and \(\text{S}_n = 2,160.\) In this case, we are asked to determine \(n.\)
$$\begin{aligned} \text{S}_{n} & = \dfrac{n}{2}(2a+(n-1)b) \\ 2,160 & = \dfrac{n}{2}(2(60) + (n-1) \cdot 12) \\ 2,160 & = n(54 + 6n) \\ n^2+9n-360 & = 0 \\ (n+24)(n-15) & = 0 \\ n=-24~\text{or} & ~n=15. \end{aligned}$$Since \(n\) cannot be negative, \(n = 15.\) Therefore, the factory worker can produce 2,160 roof tiles in 15 days.

Since the monthly installment increases continuously by Rp20,000.00, this is a contextual problem involving an arithmetic series.
Given \(\text{U}_1 = a = 250,000\) and \(b = 20,000,\) and \(\text{S}_n = 8,800,000.\) In this case, we are asked to determine \(n.\)
$$\begin{aligned} \text{S}_{n} & = \dfrac{n}{2}(2a+(n-1)b) \\ 8,800,000 & = \dfrac{n}{2}(2(250,000) + (n-1) \cdot 20,000) \\ 8,800,000 & = n(10,000n+240,000) \\ n^2+24n-880 & = 0 \\ (n+44)(n-20) & = 0 \\ n=-44~\text{or} & ~n=20. \end{aligned}$$Since \(n\) cannot be negative, \(n = 20.\) Therefore, the debt will be fully paid off in 20 months.

Since the increase in the amount saved each month is constant, this is a contextual problem involving an arithmetic series.
Sukardi’s savings in the first month are \(\text{U}_1 = 80,000\) with common difference \(b = 1,500.\) Meanwhile, Lili’s savings in the first month are \(\text{V}_1 = 100,000\) with common difference \(c = 1,000.\) Suppose their cumulative savings become exactly equal in the \(n\)-th month, namely \(\text{S}_n.\) In this case, we are asked to determine \(n.\)
$$\begin{aligned} \text{S}_n & = \text{S}_n \\ \cancel{\dfrac{n}{2}}(2\text{U}_1 + (n-1) \cdot b) & = \cancel{\dfrac{n}{2}}(2\text{V}_1 + (n-1) \cdot c) \\ 2(80,000) + (n-1) \cdot 1,500 & = 2(100,000) + (n-1) \cdot 1,000 \\ 500(n-1) & = 40,000 \\ n-1 & = 80 \\ n & = 81. \end{aligned}$$Therefore, the total amount of their savings will be exactly the same in the \(\boxed{81^\text{st}}\) month.