CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Model Questions - 4 Marks

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SECTION D — Case-Based/Source-Based Integrated Questions [4 Marks Each]

Q1. Case Study: Savings Scheme
India Post runs recurring deposit schemes to encourage people to save money. A person opens an account and deposits ₹200 in the first month, ₹250 in the second month, ₹300 in the third month, and continues to increase the deposit by ₹50 every month.

1 Marks Sub-question (a): Write the AP representing the monthly deposits for the first 4 months.
Answer:
1. Deposit in the 1st month ($a_1$) = ₹200

2. Deposit in the 2nd month ($a_2$) = ₹250

3. Deposit in the 3rd month ($a_3$) = ₹300

4. Deposit in the 4th month ($a_4$) = $300 + 50$ = ₹350

Result: Therefore, the AP representing the monthly deposits for the first 4 months is 200, 250, 300, 350.

1 Marks Sub-question (b): Find the amount deposited by the person in the 12th month.
Answer:
1. From the given AP, the first term ($a$) = 200 and the common difference ($d$) = 50.

2. We need to find the amount deposited in the 12th month ($a_{12}$). Using the AP formula $a_n = a + (n - 1)d$:
    $a_{12} = 200 + (12 - 1) \times 50$

3. Simplifying the expression:
    $a_{12} = 200 + 11 \times 50$
    $a_{12} = 200 + 550 = 750$

Result: Therefore, the amount deposited in the 12th month is ₹750.

2 Marks Sub-question (c): Calculate the total amount deposited by the person in 2 years (24 months).
Answer:
1. We need to find the sum of deposits for 24 months ($S_{24}$). Here, $a = 200$, $d = 50$, and $n = 24$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{24} = \frac{24}{2}[2(200) + (24 - 1)50]$

3. Simplifying the expression inside the brackets:
    $S_{24} = 12[400 + 23 \times 50]$
    $S_{24} = 12[400 + 1150]$

4. Adding and multiplying:
    $S_{24} = 12 \times 1550 = 18600$

Result: Therefore, the total amount deposited by the person in 2 years is ₹18,600.
Q2. Case Study: Auditorium Seating Arrangement
An auditorium has 25 rows of seats. The first row contains 20 seats, the second row contains 23 seats, the third row contains 26 seats, and so on.

1 Marks Sub-question (a): What is the common difference of the AP formed by the number of seats in each row?
Answer:
1. The number of seats in the rows form an AP: 20, 23, 26, ...

2. First term ($a_1$) = 20, Second term ($a_2$) = 23, Third term ($a_3$) = 26.

3. Common difference ($d$) = $a_2 - a_1 = 23 - 20 = 3$

Result: Therefore, the common difference of the AP is 3.

1 Marks Sub-question (b): Find the number of seats in the 15th row.
Answer:
1. Here, the first term ($a$) = 20 and the common difference ($d$) = 3.

2. We need to find the number of seats in the 15th row ($a_{15}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{15} = 20 + (15 - 1) \times 3$

3. Simplifying the expression:
    $a_{15} = 20 + 14 \times 3$
    $a_{15} = 20 + 42 = 62$

Result: Therefore, the number of seats in the 15th row is 62.

2 Marks Sub-question (c): If all seats are filled, find the total capacity of the auditorium.
Answer:
1. The total number of rows is $n = 25$. We need to find the total sum of seats in all 25 rows ($S_{25}$). Here, $a = 20$ and $d = 3$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{25} = \frac{25}{2}[2(20) + (25 - 1)3]$

3. Simplifying the expression inside the brackets:
    $S_{25} = \frac{25}{2}[40 + 24 \times 3]$
    $S_{25} = \frac{25}{2}[40 + 72]$

4. Adding and multiplying:
    $S_{25} = \frac{25}{2} \times 112$
    $S_{25} = 25 \times 56 = 1400$

Result: Therefore, the total capacity of the auditorium is 1400 seats.
Q3. Case Study: Production of Sports Equipment
A sports factory manufactures cricket bats. Due to a high demand, the factory increases its production uniformly by 50 bats every month. It produced 1,000 bats in the 2nd month.

1 Marks Sub-question (a): Find the production of bats in the 1st month.
Answer:
1. The monthly production forms an AP with a common difference ($d$) = 50.

2. We are given that the production in the 2nd month ($a_2$) = 1000.

3. Using the formula $a_2 = a + d$, where $a$ is the production in the 1st month:
    $1000 = a + 50$
    $a = 1000 - 50 = 950$

Result: Therefore, the production of bats in the 1st month is 950.

1 Marks Sub-question (b): Find the production in the 10th month.
Answer:
1. Here, the first term ($a$) = 950 and the common difference ($d$) = 50.

2. We need to find the production in the 10th month ($a_{10}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{10} = 950 + (10 - 1) \times 50$

3. Simplifying the expression:
    $a_{10} = 950 + 9 \times 50$
    $a_{10} = 950 + 450 = 1400$

Result: Therefore, the production in the 10th month is 1,400 bats.

2 Marks Sub-question (c): Find the total production of bats in the first year (12 months).
Answer:
1. We need to find the total sum of production for 12 months ($S_{12}$). Here, $a = 950$, $d = 50$, and $n = 12$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{12} = \frac{12}{2}[2(950) + (12 - 1)50]$

3. Simplifying the expression inside the brackets:
    $S_{12} = 6[1900 + 11 \times 50]$
    $S_{12} = 6[1900 + 550]$

4. Adding and multiplying:
    $S_{12} = 6 \times 2450 = 14700$

Result: Therefore, the total production of bats in the first year is 14,700 bats.
Q4. Case Study: Flower Bed Design
A gardener is planting flowering plants in circular rings. In the first flower ring, there are 23 rose plants, in the second ring there are 21, in the third there are 19, and so on. There are 5 plants in the last ring.

1 Marks Sub-question (a): Is the number of plants in each ring forming an AP? Justify with common difference.
Answer:
1. The number of plants in the rings form the sequence: 23, 21, 19, ...

2. First term ($a_1$) = 23, Second term ($a_2$) = 21, Third term ($a_3$) = 19.

3. Finding the differences between consecutive terms:
    $a_2 - a_1 = 21 - 23 = -2$
    $a_3 - a_2 = 19 - 21 = -2$

4. Since the difference between any term and its preceding term is constant ($d = -2$), the sequence forms an AP.

Result: Yes, it forms an AP with a common difference of $-2$.

1 Marks Sub-question (b): Find the total number of rings formed in the garden.
Answer:
1. Here, the first term ($a$) = 23, common difference ($d$) = $-2$, and the last term ($a_n$) = 5.

2. Using the AP formula $a_n = a + (n - 1)d$:
    $5 = 23 + (n - 1)(-2)$

3. Solving for $n$:
    $5 - 23 = -2(n - 1)$
    $-18 = -2(n - 1)$
    $n - 1 = \frac{-18}{-2} = 9$
    $n = 9 + 1 = 10$

Result: Therefore, the total number of rings formed in the garden is 10.

2 Marks Sub-question (c): Find the total number of rose plants planted across all the rings.
Answer:
1. We need to find the total sum of plants in all 10 rings ($S_{10}$). Here, $a = 23$, $l = 5$, and $n = 10$.

2. Using the AP sum formula when the last term is known ($S_n = \frac{n}{2}(a + l)$):
    $S_{10} = \frac{10}{2}(23 + 5)$

3. Simplifying the expression:
    $S_{10} = 5 \times 28$
    $S_{10} = 140$

Result: Therefore, the total number of rose plants planted across all the rings is 140.
Q5. Case Study: The Ladder Design
A wooden ladder has rungs that decrease uniformly in length from bottom to top. The bottom rung is $45\text{ cm}$ long and the top rung is $25\text{ cm}$ long. The rungs are spaced $25\text{ cm}$ apart, and the total height of the ladder is $2.5\text{ m}$ ($250\text{ cm}$).

1 Marks Sub-question (a): Calculate the total number of rungs on the ladder.
Answer:
1. Total height of the ladder = $250\text{ cm}$ and the rungs are spaced $25\text{ cm}$ apart.

2. The number of intervals between rungs = $\frac{250}{25} = 10$.

3. Since there is a rung at both the bottom and the top ends, the total number of rungs ($n$) is the number of intervals plus $1$.

    $n = 10 + 1 = 11$

Result: Therefore, the total number of rungs on the ladder is 11.

1 Marks Sub-question (b): Find the common difference in the lengths of consecutive rungs.
Answer:
1. Here, the first term ($a$) = $45\text{ cm}$, the last term ($a_{11}$) = $25\text{ cm}$, and $n = 11$.

2. Using the AP formula $a_n = a + (n - 1)d$:
    $25 = 45 + (11 - 1)d$

3. Solving for $d$:
    $25 - 45 = 10d$
    $-20 = 10d$
    $d = \frac{-20}{10} = -2$

Result: Therefore, the common difference in the lengths of consecutive rungs is $-2\text{ cm}$.

2 Marks Sub-question (c): Find the total length of the wood required to make all the rungs.
Answer:
1. We need to find the sum of the lengths of all 11 rungs ($S_{11}$). Here, $a = 45$, $l = 25$, and $n = 11$.

2. Using the AP sum formula when the last term is known ($S_n = \frac{n}{2}(a + l)$):
    $S_{11} = \frac{11}{2}(45 + 25)$

3. Simplifying the expression:
    $S_{11} = \frac{11}{2} \times 70$
    $S_{11} = 11 \times 35 = 385$

Result: Therefore, the total length of the wood required to make all the rungs is $385\text{ cm}$ (or $3.85\text{ m}$).
Q6. Case Study: Tree Plantation Drive
To reduce air pollution, students of a school decide to plant trees in and around their campus. It was decided that the number of trees that each section of each class will plant corresponds to the class number (e.g., Class 1 will plant 1 tree, Class 2 will plant 2 trees, up to Class 12). Each class has 3 sections.

1 Marks Sub-question (a): How many trees will be planted by all sections of Class 5?
Answer:
1. Each section of Class 5 plants $5$ trees.

2. Since there are $3$ sections in Class 5, the total number of trees planted by all sections of Class 5 = $5 \times 3 = 15$.

Result: Therefore, 15 trees will be planted by all sections of Class 5.

1 Marks Sub-question (b): Write the AP representing total trees planted by each class from Class 1 to Class 12.
Answer:
1. Trees planted by Class 1 (3 sections) = $1 \times 3 = 3$

2. Trees planted by Class 2 (3 sections) = $2 \times 3 = 6$

3. Trees planted by Class 3 (3 sections) = $3 \times 3 = 9$

4. Continuing this pattern up to Class 12, the sequence of total trees planted by each class is $3, 6, 9, \dots, 36$.

Result: Therefore, the AP representing the total trees planted by each class is 3, 6, 9, ..., 36.

2 Marks Sub-question (c): Find the total number of trees planted by the students of the whole school from Class 1 to Class 12.
Answer:
1. We need to find the sum of the AP for 12 classes ($S_{12}$). Here, first term ($a$) = $3$, common difference ($d$) = $3$, last term ($l$) = $36$, and $n = 12$.

2. Using the AP sum formula when the last term is known ($S_n = \frac{n}{2}(a + l)$):
    $S_{12} = \frac{12}{2}(3 + 36)$

3. Simplifying the expression:
    $S_{12} = 6 \times 39$
    $S_{12} = 234$

Result: Therefore, the total number of trees planted by the students of the whole school is 234.
Q7. Case Study: Car Loan Repayment
A person takes a car loan from a bank and agrees to repay a total loan of ₹1,18,000 in monthly installments. He pays the first installment of ₹1,000 and increases each subsequent installment by ₹100.

1 Marks Sub-question (a): Write the first 3 terms of the AP formed by the monthly installments.
Answer:
1. First installment ($a_1$) = ₹1,000

2. Second installment ($a_2$) = $1000 + 100$ = ₹1,100

3. Third installment ($a_3$) = $1100 + 100$ = ₹1,200

Result: Therefore, the first 3 terms of the AP are ₹1,000, ₹1,100, ₹1,200.

1 Marks Sub-question (b): Calculate the amount paid by him in the 30th installment.
Answer:
1. Here, the first term ($a$) = 1000 and the common difference ($d$) = 100.

2. We need to find the amount paid in the 30th installment ($a_{30}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{30} = 1000 + (30 - 1) \times 100$

3. Simplifying the expression:
    $a_{30} = 1000 + 29 \times 100$
    $a_{30} = 1000 + 2900 = 3900$

Result: Therefore, the amount paid in the 30th installment is ₹3,900.

2 Marks Sub-question (c): Find the total amount he has paid after 30 installments.
Answer:
1. We need to find the sum of installments for 30 months ($S_{30}$). Here, $a = 1000$, $d = 100$, and $n = 30$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{30} = \frac{30}{2}[2(1000) + (30 - 1)100]$

3. Simplifying the expression inside the brackets:
    $S_{30} = 15[2000 + 29 \times 100]$
    $S_{30} = 15[2000 + 2900]$

4. Adding and multiplying:
    $S_{30} = 15 \times 4900 = 73500$

Result: Therefore, the total amount paid after 30 installments is ₹73,500.
Q8. Case Study: The Running Track Practice
An athlete is preparing for a marathon. On the first day of training, she runs $2\text{ km}$. Each subsequent day, she increases her running distance by $0.5\text{ km}$.

1 Marks Sub-question (a): Find the distance she runs on the 10th day of her training.
Answer:
1. Here, the first term ($a$) = $2\text{ km}$ and the common difference ($d$) = $0.5\text{ km}$.

2. We need to find the distance run on the 10th day ($a_{10}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{10} = 2 + (10 - 1) \times 0.5$

3. Simplifying the expression:
    $a_{10} = 2 + 9 \times 0.5$
    $a_{10} = 2 + 4.5 = 6.5$

Result: Therefore, the distance she runs on the 10th day is $6.5\text{ km}$.

1 Marks Sub-question (b): On which day will her daily running distance reach $10\text{ km}$?
Answer:
1. Here, $a = 2$, $d = 0.5$, and the target distance ($a_n$) = $10\text{ km}$.

2. Using the formula $a_n = a + (n - 1)d$:
    $10 = 2 + (n - 1) \times 0.5$

3. Solving for $n$:
    $10 - 2 = 0.5(n - 1)$
    $8 = 0.5(n - 1)$
    $n - 1 = \frac{8}{0.5} = 16$
    $n = 16 + 1 = 17$

Result: Therefore, her daily running distance will reach $10\text{ km}$ on the 17th day.

2 Marks Sub-question (c): Find the total distance she covers in 15 days of training.
Answer:
1. We need to find the sum of distances for 15 days ($S_{15}$). Here, $a = 2$, $d = 0.5$, and $n = 15$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{15} = \frac{15}{2}[2(2) + (15 - 1) \times 0.5]$

3. Simplifying the expression inside the brackets:
    $S_{15} = \frac{15}{2}[4 + 14 \times 0.5]$
    $S_{15} = \frac{15}{2}[4 + 7]$
    $S_{15} = \frac{15}{2} \times 11$

4. Calculating the final value:
    $S_{15} = \frac{165}{2} = 82.5$

Result: Therefore, the total distance she covers in 15 days of training is $82.5\text{ km}$.
Q9. Case Study: Football Ground Construction
The management of a stadium is building steps for a spectator stand. Each step is built out of solid concrete and rises by a height of $15\text{ cm}$, has a width of $50\text{ cm}$, and a length of $50\text{ m}$. The total number of steps is 15.

1 Marks Sub-question (a): Find the cross-sectional area of the first step.
Answer:
1. Height of the step = $15\text{ cm} = 0.15\text{ m}$ (or keep in cm: width = $50\text{ cm}$, height = $15\text{ cm}$).

2. The cross-sectional area of a step is given by width $\times$ height:
    Area = $50\text{ cm} \times 15\text{ cm} = 750\text{ cm}^2$
    (In meters: $0.5\text{ m} \times 0.15\text{ m} = 0.075\text{ m}^2$)

Result: Therefore, the cross-sectional area of the first step is $750\text{ cm}^2$ (or $0.075\text{ m}^2$).

1 Marks Sub-question (b): Find the volume of concrete required for the 3rd step.
Answer:
1. Length of each step ($l$) = $50\text{ m} = 5000\text{ cm}$.

2. Volume of a step = Cross-sectional area $\times$ Length.
    Volume of the 1st step ($a_1$) = $0.075\text{ m}^2 \times 50\text{ m} = 3.75\text{ m}^3$.

3. The volume of concrete required for each subsequent step increases uniformly because each step builds on top of the previous height pattern, forming an AP where the volume of each step increases by the volume of one base step ($d = 3.75\text{ m}^3$).
    Volume of 3rd step ($a_3$) = $a + 2d = 3.75 + 2(3.75) = 3(3.75) = 11.25\text{ m}^3$.

Result: Therefore, the volume of concrete required for the 3rd step is $11.25\text{ m}^3$.

2 Marks Sub-question (c): Calculate the total volume of concrete required to build all 15 steps.
Answer:
1. We need to find the sum of volumes for all 15 steps ($S_{15}$). Here, the first term ($a$) = $3.75\text{ m}^3$, common difference ($d$) = $3.75\text{ m}^3$, and $n = 15$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{15} = \frac{15}{2}[2(3.75) + (15 - 1) \times 3.75]$

3. Simplifying the expression inside the brackets:
    $S_{15} = \frac{15}{2}[7.5 + 14 \times 3.75]$
    $S_{15} = \frac{15}{2}[7.5 + 52.5]$
    $S_{15} = \frac{15}{2} \times 60$

4. Calculating the final value:
    $S_{15} = 15 \times 30 = 450$

Result: Therefore, the total volume of concrete required to build all 15 steps is $450\text{ m}^3$.
Q10. Case Study: Designing a Spiral
A spiral is made up of successive semicircles, with centers alternately at $A$ and $B$, starting with a center at $A$, of radii $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \dots$ as shown in standard mathematical models.

1 Marks Sub-question (a): Write the lengths of the first three semicircles using $\pi = \frac{22}{7}$.
Answer:
1. The length of a semicircle is given by the formula $l = \pi r$.

2. For the first semicircle with radius $r_1 = 0.5\text{ cm}$:
    $l_1 = \pi \times 0.5 = \frac{22}{7} \times \frac{1}{2} = \frac{11}{7}\text{ cm}$

3. For the second semicircle with radius $r_2 = 1.0\text{ cm}$:
    $l_2 = \pi \times 1.0 = \frac{22}{7} \times 1 = \frac{22}{7}\text{ cm}$

4. For the third semicircle with radius $r_3 = 1.5\text{ cm}$:
    $l_3 = \pi \times 1.5 = \frac{22}{7} \times \frac{3}{2} = \frac{33}{7}\text{ cm}$

Result: Therefore, the lengths of the first three semicircles are $\frac{11}{7}\text{ cm}$, $\frac{22}{7}\text{ cm}$, and $\frac{33}{7}\text{ cm}$.

1 Marks Sub-question (b): Show that these lengths form an AP and find its common difference.
Answer:
1. The sequence of lengths is $\frac{11}{7}, \frac{22}{7}, \frac{33}{7}, \dots$

2. Finding the differences between consecutive terms:
    $l_2 - l_1 = \frac{22}{7} - \frac{11}{7} = \frac{11}{7}$
    $l_3 - l_2 = \frac{33}{7} - \frac{22}{7} = \frac{11}{7}$

3. Since the difference between any term and its preceding term is constant ($d = \frac{11}{7}$), the sequence forms an AP.

Result: Yes, it forms an AP with a common difference of $\frac{11}{7}\text{ cm}$.

2 Marks Sub-question (c): Find the total length of such a spiral made up of 13 consecutive semicircles.
Answer:
1. We need to find the sum of the lengths of 13 semicircles ($S_{13}$). Here, first term ($a$) = $\frac{11}{7}$, common difference ($d$) = $\frac{11}{7}$, and $n = 13$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{13} = \frac{13}{2}\left[2\left(\frac{11}{7}\right) + (13 - 1)\left(\frac{11}{7}\right)\right]$

3. Simplifying the expression inside the brackets:
    $S_{13} = \frac{13}{2}\left[\frac{22}{7} + 12 \times \frac{11}{7}\right]$
    $S_{13} = \frac{13}{2}\left[\frac{22}{7} + \frac{132}{7}\right]$
    $S_{13} = \frac{13}{2}\left[\frac{154}{7}\right]$

4. Calculating the final value:
    $S_{13} = \frac{13}{2} \times 22 = 13 \times 11 = 143$

Result: Therefore, the total length of the spiral made up of 13 consecutive semicircles is $143\text{ cm}$.
Q11. Case Study: Daily Pocket Money Savings
Anuj gets pocket money from his father. He saves ₹2.75 on the first day, ₹3.00 on the second day, ₹3.25 on the third day, and continues this pattern.

1 Marks Sub-question (a): What is the common difference of his daily savings?
Answer:
1. The daily savings form an AP: ₹2.75, ₹3.00, ₹3.25, ...

2. First term ($a_1$) = 2.75, Second term ($a_2$) = 3.00, Third term ($a_3$) = 3.25.

3. Common difference ($d$) = $a_2 - a_1 = 3.00 - 2.75 = 0.25$

Result: Therefore, the common difference of his daily savings is ₹0.25.

1 Marks Sub-question (b): Find the amount saved by Anuj on the 14th day.
Answer:
1. Here, the first term ($a$) = 2.75 and the common difference ($d$) = 0.25.

2. We need to find the savings on the 14th day ($a_{14}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{14} = 2.75 + (14 - 1) \times 0.25$

3. Simplifying the expression:
    $a_{14} = 2.75 + 13 \times 0.25$
    $a_{14} = 2.75 + 3.25 = 6.00$

Result: Therefore, the amount saved by Anuj on the 14th day is ₹6.00.

2 Marks Sub-question (c): Calculate his total savings in 30 days.
Answer:
1. We need to find the sum of savings for 30 days ($S_{30}$). Here, $a = 2.75$, $d = 0.25$, and $n = 30$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{30} = \frac{30}{2}[2(2.75) + (30 - 1) \times 0.25]$

3. Simplifying the expression inside the brackets:
    $S_{30} = 15[5.50 + 29 \times 0.25]$
    $S_{30} = 15[5.50 + 7.25]$
    $S_{30} = 15 \times 12.75$

4. Calculating the final value:
    $S_{30} = 191.25$

Result: Therefore, his total savings in 30 days is ₹191.25.
Q12. Case Study: Candies Distribution Problem
Aditya is celebrating his birthday and distributes candy packets in school. He arranges candies in rows such that the first row has 3 candies, the second row has 5 candies, the third row has 7 candies, and so on. Total candies in the packet are 120.

1 Marks Sub-question (a): Find the number of candies placed in the 10th row.
Answer:
1. The number of candies in each row forms an AP: 3, 5, 7, ...

2. Here, the first term ($a$) = 3 and the common difference ($d$) = $5 - 3 = 2$.

3. We need to find the number of candies in the 10th row ($a_{10}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{10} = 3 + (10 - 1) \times 2$

4. Simplifying the expression:
    $a_{10} = 3 + 9 \times 2$
    $a_{10} = 3 + 18 = 21$

Result: Therefore, the number of candies placed in the 10th row is 21.

1 Marks Sub-question (b): Find the difference in the number of candies placed in the 7th and 3rd rows.
Answer:
1. We need to find the difference $a_7 - a_3$.

2. Using the formula $a_n = a + (n - 1)d$, we can write:
    $a_7 = a + 6d$
    $a_3 = a + 2d$

3. Finding the difference:
    $a_7 - a_3 = (a + 6d) - (a + 2d) = 4d$

4. Substituting the common difference $d = 2$:
    $4 \times 2 = 8$

Result: Therefore, the difference in the number of candies placed in the 7th and 3rd rows is 8.

2 Marks Sub-question (c): Find the total number of rows created using all 120 candies.
Answer:
1. Here, the total sum of candies ($S_n$) = 120, first term ($a$) = 3, and common difference ($d$) = 2.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $120 = \frac{n}{2}[2(3) + (n - 1) \times 2]$

3. Simplifying and solving for $n$:
    $120 = \frac{n}{2}[6 + 2n - 2]$
    $120 = \frac{n}{2}[2n + 4]$
    $120 = n(n + 2)$
    $n^2 + 2n - 120 = 0$

4. Factorizing the quadratic equation:
    $n^2 + 12n - 10n - 120 = 0$
    $n(n + 12) - 10(n + 12) = 0$
    $(n - 10)(n + 12) = 0$

5. Since the number of rows cannot be negative, $n = 10$.

Result: Therefore, the total number of rows created using all 120 candies is 10.
Q13. Case Study: Bus Ticketing System & Fares
A roadways bus service charges a base fare for the first kilometer and then a fixed flat rate for every additional kilometer traveled. The total fare for a 4 km journey is ₹60, and for a 10 km journey is ₹120.

1 Marks Sub-question (a): Find the fixed rate per additional kilometer (common difference).
Answer:
1. Let the base fare for the 1st km be $a$ and the fixed rate per additional km be $d$.

2. The total fare for a 4 km journey ($a_4$) is given by:
    $a_4 = a + 3d = 60$

3. The total fare for a 10 km journey ($a_{10}$) is given by:
    $a_{10} = a + 9d = 120$

4. Subtracting the first equation from the second equation:
    $(a + 9d) - (a + 3d) = 120 - 60$
    $6d = 60$
    $d = \frac{60}{6} = 10$

Result: Therefore, the fixed rate per additional kilometer is ₹10.

1 Marks Sub-question (b): Find the base fare for the first kilometer.
Answer:
1. We know that the fare for a 4 km journey is $a + 3d = 60$ and the common difference $d = 10$.

2. Substituting the value of $d$ into the equation:
    $a + 3(10) = 60$
    $a + 30 = 60$
    $a = 60 - 30 = 30$

Result: Therefore, the base fare for the first kilometer is ₹30.

2 Marks Sub-question (c): Calculate the total fare for a journey of 25 km.
Answer:
1. We need to find the total fare for a 25 km journey ($a_{25}$). Here, first term ($a$) = 30, common difference ($d$) = 10, and $n = 25$.

2. Using the AP formula $a_n = a + (n - 1)d$:
    $a_{25} = 30 + (25 - 1) \times 10$

3. Simplifying the expression:
    $a_{25} = 30 + 24 \times 10$
    $a_{25} = 30 + 240 = 270$

Result: Therefore, the total fare for a journey of 25 km is ₹270.
Q14. Case Study: Dropping Pebbles / Potato Race
In a school sports day potato race, a bucket is placed at the starting point, which is $5\text{ m}$ from the first potato, and other potatoes are placed $3\text{ m}$ apart in a straight line.

1 Marks Sub-question (a): What is the total distance run by the competitor to retrieve the 1st potato and put it back?
Answer:
1. Distance of the first potato from the bucket = $5\text{ m}$.

2. The competitor runs to the potato and brings it back to the bucket, covering the distance twice.

3. Total distance for the 1st potato ($a_1$) = $2 \times 5 = 10\text{ m}$.

Result: Therefore, the total distance run to retrieve the 1st potato is $10\text{ m}$.

1 Marks Sub-question (b): Write the AP representing the total distances run for the first 3 potatoes.
Answer:
1. Distance for the 1st potato ($a_1$) = $2 \times 5 = 10\text{ m}$.

2. Distance for the 2nd potato ($a_2$) = $2 \times (5 + 3) = 2 \times 8 = 16\text{ m}$.

3. Distance for the 3rd potato ($a_3$) = $2 \times (5 + 3 + 3) = 2 \times 11 = 22\text{ m}$.

Result: Therefore, the AP representing the total distances run for the first 3 potatoes is 10, 16, 22.

2 Marks Sub-question (c): Find the total distance the competitor has to run to complete all 10 potatoes.
Answer:
1. We need to find the sum of distances for 10 potatoes ($S_{10}$). Here, first term ($a$) = 10, common difference ($d$) = $16 - 10 = 6$, and $n = 10$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{10} = \frac{10}{2}[2(10) + (10 - 1) \times 6]$

3. Simplifying the expression inside the brackets:
    $S_{10} = 5[20 + 9 \times 6]$
    $S_{10} = 5[20 + 54]$
    $S_{10} = 5 \times 74$

4. Calculating the final value:
    $S_{10} = 370$

Result: Therefore, the total distance the competitor has to run to complete all 10 potatoes is $370\text{ m}$.
Q15. Case Study: Target Archery Practice
An archer shoots arrows at a target board. He scores 10 points for hitting the bullseye, but his score reduces uniformly by 2 points for every concentric ring outwards (8, 6, 4...). Suppose a trainee hits rings such that her scores form a descending AP.

1 Marks Sub-question (a): If her 4th ring score is 4 and 8th ring score is $-4$, find her first ring score.
Answer:
1. Let the first ring score be $a$ and the common difference be $d$.

2. We are given the 4th score ($a_4 = 4$) and the 8th score ($a_8 = -4$).
    $a_4 = a + 3d = 4$
    $a_8 = a + 7d = -4$

3. Subtracting the first equation from the second equation:
    $(a + 7d) - (a + 3d) = -4 - 4$
    $4d = -8$
    $d = -2$

4. Substituting $d = -2$ into the first equation ($a + 3(-2) = 4$):
    $a - 6 = 4$
    $a = 4 + 6 = 10$

Result: Therefore, her first ring score is 10.

1 Marks Sub-question (b): Find the common difference of this score sequence.
Answer:
1. From part (a), we set up the linear equations using $a_4$ and $a_8$ to find the common difference ($d$).

2. $4d = -8 \implies d = -2$.

Result: Therefore, the common difference of this score sequence is $-2$.

2 Marks Sub-question (c): Find the sum of her scores for the first 10 rings hit consecutively.
Answer:
1. We need to find the sum of the first 10 scores ($S_{10}$). Here, $a = 10$, $d = -2$, and $n = 10$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{10} = \frac{10}{2}[2(10) + (10 - 1)(-2)]$

3. Simplifying the expression inside the brackets:
    $S_{10} = 5[20 + 9(-2)]$
    $S_{10} = 5[20 - 18]$
    $S_{10} = 5[2]$

4. Calculating the final value:
    $S_{10} = 10$

Result: Therefore, the sum of her scores for the first 10 rings hit consecutively is 10.
Q16. Case Study: Manufacturing Arch Bars
A steel factory manufactures metal rods. The length of the rods produced declines due to machine wear and tear by a fixed margin every hour. The 3rd hour production length is $400\text{ mm}$ and the 7th hour production length is $320\text{ mm}$.

1 Marks Sub-question (a): Find the fixed reduction value per hour.
Answer:
1. Let the length in the 1st hour be $a$ and the fixed reduction per hour (common difference) be $d$.

2. We are given the 3rd hour length ($a_3 = 400$) and the 7th hour length ($a_7 = 320$).
    $a_3 = a + 2d = 400$
    $a_7 = a + 6d = 320$

3. Subtracting the first equation from the second equation:
    $(a + 6d) - (a + 2d) = 320 - 400$
    $4d = -80$
    $d = \frac{-80}{4} = -20$

Result: Therefore, the fixed reduction value per hour is $-20\text{ mm}$.

1 Marks Sub-question (b): Find the length of the rod produced in the 1st hour.
Answer:
1. We know that the length for the 3rd hour is $a + 2d = 400$ and the common difference $d = -20$.

2. Substituting the value of $d$ into the equation:
    $a + 2(-20) = 400$
    $a - 40 = 400$
    $a = 400 + 40 = 440$

Result: Therefore, the length of the rod produced in the 1st hour is $440\text{ mm}$.

2 Marks Sub-question (c): Find the total length of all rods produced cumulatively in the first 10 hours.
Answer:
1. We need to find the sum of lengths for the first 10 hours ($S_{10}$). Here, $a = 440$, $d = -20$, and $n = 10$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{10} = \frac{10}{2}[2(440) + (10 - 1)(-20)]$

3. Simplifying the expression inside the brackets:
    $S_{10} = 5[880 + 9(-20)]$
    $S_{10} = 5[880 - 180]$
    $S_{10} = 5 \times 700$

4. Calculating the final value:
    $S_{10} = 3500$

Result: Therefore, the total length of all rods produced cumulatively in the first 10 hours is $3500\text{ mm}$ (or $3.5\text{ m}$).
Q17. Case Study: Architectural Staircase Design
An architect is designing a grand staircase for a hotel lobby. The vertical height of each step is uniform. The height of the 4th step from the floor is $60\text{ cm}$ and the height of the 11th step is $165\text{ cm}$.

1 Marks Sub-question (a): Find the height of each individual step.
Answer:
1. Let the height of the 1st step be $a$ and the uniform height of each individual step (common difference) be $d$.

2. We are given the height of the 4th step ($a_4 = 60$) and the 11th step ($a_{11} = 165$).
    $a_4 = a + 3d = 60$
    $a_{11} = a + 10d = 165$

3. Subtracting the first equation from the second equation:
    $(a + 10d) - (a + 3d) = 165 - 60$
    $7d = 105$
    $d = \frac{105}{7} = 15$

Result: Therefore, the height of each individual step is $15\text{ cm}$.

1 Marks Sub-question (b): Find the height of the 20th step from the floor.
Answer:
1. First, find the height of the 1st step ($a$) by substituting $d = 15$ into $a + 3d = 60$:
    $a + 3(15) = 60$
    $a + 45 = 60 \implies a = 15$

2. We need to find the height of the 20th step ($a_{20}$). Using the formula $a_n = a + (n - 1)d$:
    $a_{20} = 15 + (20 - 1) \times 15$

3. Simplifying the expression:
    $a_{20} = 15 + 19 \times 15$
    $a_{20} = 15 + 285 = 300$

Result: Therefore, the height of the 20th step from the floor is $300\text{ cm}$ (or $3\text{ m}$).

2 Marks Sub-question (c): If there are 30 steps in total, find the total height of the top platform from the ground.
Answer:
1. The height of the top platform is given by the height of the 30th step ($a_{30}$). Here, $a = 15$, $d = 15$, and $n = 30$.

2. Using the formula $a_n = a + (n - 1)d$ for $n = 30$:
    $a_{30} = 15 + (30 - 1) \times 15$

3. Simplifying the expression:
    $a_{30} = 15 + 29 \times 15$
    $a_{30} = 15 + 435 = 450$

Result: Therefore, the total height of the top platform from the ground is $450\text{ cm}$ (or $4.5\text{ m}$).
Q18. Case Study: Cell Phone Plan Billing
A telecom company offers a data pack plan where a customer uses fixed MBs initially, and then consumes extra data daily that increases in an arithmetic progression. On day 2, the user consumed $150\text{ MB}$, and on day 5, $240\text{ MB}$.

1 Marks Sub-question (a): Find the consumption value on day 1.
Answer:
1. Let the consumption on day 1 be $a$ and the daily increase (common difference) be $d$.

2. We are given the consumption on day 2 ($a_2 = 150$) and day 5 ($a_5 = 240$).
    $a_2 = a + d = 150$
    $a_5 = a + 4d = 240$

3. Subtracting the first equation from the second equation:
    $(a + 4d) - (a + d) = 240 - 150$
    $3d = 90 \implies d = 30$

4. Substituting $d = 30$ into the first equation ($a + 30 = 150$):
    $a = 150 - 30 = 120$

Result: Therefore, the consumption value on day 1 is $120\text{ MB}$.

1 Marks Sub-question (b): Find the daily fixed increase in data consumption.
Answer:
1. From part (a), using the linear equations for day 2 and day 5, we solved for $d$.

2. $3d = 90 \implies d = 30$.

Result: Therefore, the daily fixed increase in data consumption is $30\text{ MB}$.

2 Marks Sub-question (c): Calculate the total data consumed in the first 10 days.
Answer:
1. We need to find the sum of data consumed for the first 10 days ($S_{10}$). Here, $a = 120$, $d = 30$, and $n = 10$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{10} = \frac{10}{2}[2(120) + (10 - 1) \times 30]$

3. Simplifying the expression inside the brackets:
    $S_{10} = 5[240 + 9 \times 30]$
    $S_{10} = 5[240 + 270]$
    $S_{10} = 5 \times 510$

4. Calculating the final value:
    $S_{10} = 2550$

Result: Therefore, the total data consumed in the first 10 days is $2550\text{ MB}$ (or $2.55\text{ GB}$).
Q19. Case Study: Exponential vs Linear Growth (Salary Increment)
An employee joins an IT firm. His starting annual salary package is ₹3,00,000, and he gets an annual increment of ₹20,000 every year consistently.

1 Marks Sub-question (a): What will be his salary in the 5th year of service?
Answer:
1. The annual salaries form an AP where the first term ($a$) = 3,00,000 and the common difference ($d$) = 20,000.

2. We need to find the salary in the 5th year ($a_5$). Using the formula $a_n = a + (n - 1)d$:
    $a_5 = 3,00,000 + (5 - 1) \times 20,000$

3. Simplifying the expression:
    $a_5 = 3,00,000 + 4 \times 20,000$
    $a_5 = 3,00,000 + 80,000 = 3,80,000$

Result: Therefore, his salary in the 5th year of service is ₹3,80,000.

1 Marks Sub-question (b): In which year will his annual salary cross ₹5,00,000?
Answer:
1. We need to find the year $n$ when $a_n > 5,00,000$, using $a = 3,00,000$ and $d = 20,000$.

2. Setting up the inequality/equation:
    $3,00,000 + (n - 1) \times 20,000 = 5,00,000$

3. Solving for $n$:
    $(n - 1) \times 20,000 = 5,00,000 - 3,00,000$
    $(n - 1) \times 20,000 = 2,00,000$
    $n - 1 = \frac{2,00,000}{20,000} = 10$
    $n = 10 + 1 = 11$

Result: Therefore, his annual salary will cross ₹5,00,000 in the 11th year.

2 Marks Sub-question (c): Calculate his total cumulative earnings over a period of 10 years.
Answer:
1. We need to find the sum of earnings for the first 10 years ($S_{10}$). Here, $a = 3,00,000$, $d = 20,000$, and $n = 10$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{10} = \frac{10}{2}[2(3,00,000) + (10 - 1) \times 20,000]$

3. Simplifying the expression inside the brackets:
    $S_{10} = 5[6,00,000 + 9 \times 20,000]$
    $S_{10} = 5[6,00,000 + 1,80,000]$
    $S_{10} = 5 \times 7,80,000$

4. Calculating the final value:
    $S_{10} = 39,00,000$

Result: Therefore, his total cumulative earnings over a period of 10 years is ₹39,00,000.
Q20. Case Study: Water Reservoir Management
A water tank is filled using pumps. Due to leakage and variable flow, the water collected everyday follows an AP. On day 3, the tank recorded $500\text{ liters}$ of water inflow, and on day 7, it recorded $700\text{ liters}$.

1 Marks Sub-question (a): Find the water inflow on the 1st day.
Answer:
1. Let the water inflow on the 1st day be $a$ and the common difference be $d$.

2. We are given the inflow on day 3 ($a_3 = 500$) and day 7 ($a_7 = 700$).
    $a_3 = a + 2d = 500$
    $a_7 = a + 6d = 700$

3. Subtracting the first equation from the second equation:
    $(a + 6d) - (a + 2d) = 700 - 500$
    $4d = 200 \implies d = 50$

4. Substituting $d = 50$ into the first equation ($a + 2(50) = 500$):
    $a + 100 = 500 \implies a = 400$

Result: Therefore, the water inflow on the 1st day is $400\text{ liters}$.

1 Marks Sub-question (b): Find the common difference in daily water inflow.
Answer:
1. From part (a), using the linear equations for day 3 and day 7, we solved for $d$.

2. $4d = 200 \implies d = 50$.

Result: Therefore, the common difference in daily water inflow is $50\text{ liters}$.

2 Marks Sub-question (c): Calculate the total volume of water collected in the reservoir in the first 15 days.
Answer:
1. We need to find the sum of water inflow for the first 15 days ($S_{15}$). Here, $a = 400$, $d = 50$, and $n = 15$.

2. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
    $S_{15} = \frac{15}{2}[2(400) + (15 - 1) \times 50]$

3. Simplifying the expression inside the brackets:
    $S_{15} = \frac{15}{2}[800 + 14 \times 50]$
    $S_{15} = \frac{15}{2}[800 + 700]$
    $S_{15} = \frac{15}{2} \times 1500$

4. Calculating the final value:
    $S_{15} = 15 \times 750 = 11,250$

Result: Therefore, the total volume of water collected in the reservoir in the first 15 days is $11,250\text{ liters}$.

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