Chapter 5 : Class 10 : Arithmetic Progressions: Case Study Question Bank

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Chapter 5: Arithmetic Progressions

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Carbon Offset 4 Marks

Carbon Offset Plantation

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A reforestation campaign plants 100 saplings in Year 1 and increases the count by 20 saplings every subsequent year:

• First term (a): 100 saplings
• Common difference (d): 20 saplings/year
• Progression: 100, 120, 140, 160, ...

Q1. Conceptual Understanding 1 Mark

Calculate the exact number of carbon-absorbing saplings the organization will plant in the 10th year of their campaign.

(A) 260 saplings

(B) 280 saplings

(C) 300 saplings

(D) 320 saplings

Q2. Application 1 Mark

In which specific campaign year will the organization plant exactly 500 saplings?

(A) 19th year

(B) 20th year

(C) 21st year

(D) 22nd year

Q3. Analytical Reasoning 2 Marks

Find the cumulative total number of carbon-absorbing saplings planted over the first 12 years of the project.

(A) 2,400 saplings

(B) 2,520 saplings

(C) 2,640 saplings

(D) 2,800 saplings

OR (Alternative Q3)

If the organization sets an ambitious long-term target to plant a cumulative total of 3,600 saplings, how many years are required?

(A) 15 years

(B) 16 years

(C) 18 years

(D) 20 years

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a = 100, d = 20. a₁₀ = a + 9d = 100 + 9(20) = 100 + 180 = 280 saplings.	[0.5 Mark] for a and d values. [0.5 Mark] for calculating 280.
Q2	a_n = 100 + (n − 1)20 = 500 ⇒ (n − 1)20 = 400 ⇒ n − 1 = 20 ⇒ n = 21.	[0.5 Mark] for setting up a_n equation. [0.5 Mark] for evaluating n = 21.
Q3	S₁₂ = (12/2)[2(100) + (12 − 1)20] = 6[200 + 220] = 6(420) = 2,520 saplings.	[1 Mark] for S_n substitution. [1 Mark] for final answer 2,520.
Q3 (OR)	(n/2)[2(100) + (n − 1)20] = 3600 ⇒ n[10n + 90] = 3600 ⇒ n² + 9n − 360 = 0. (n − 15)(n + 24) = 0 ⇒ n = 15 years (rejecting −24).	[1 Mark] for quadratic formulation. [1 Mark] for solving n = 15 years.

Case Study 2 Aerospace Constellation 4 Marks

LEO Satellite Constellation

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A private aerospace tech firm deploys a satellite constellation in concentric rings:

• Ring 1 (a): 10 satellites
• Increase per ring (d): 4 satellites
• Distribution: 10, 14, 18, 22, ...

Q1. Conceptual Understanding 1 Mark

Calculate the number of satellites that must be deployed in the 12th orbital ring.

(A) 54 satellites

(B) 50 satellites

(C) 58 satellites

(D) 48 satellites

Q2. Application 1 Mark

If an outer ring contains exactly 90 satellites, determine the ring number of this boundary.

(A) Ring 20

(B) Ring 21

(C) Ring 22

(D) Ring 24

Q3. Analytical Reasoning 2 Marks

Find the total number of satellites deployed in the constellation across the first 15 orbital rings.

(A) 540 satellites

(B) 550 satellites

(C) 600 satellites

(D) 570 satellites

OR (Alternative Q3)

If maximum operational limit is 960 satellites, calculate maximum number of rings that can be fully populated.

(A) 20 rings

(B) 22 rings

(C) 24 rings

(D) 18 rings

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁₂ = 10 + 11(4) = 10 + 44 = 54 satellites.	[1 Mark] for evaluating a₁₂ = 54.
Q2	10 + (n − 1)4 = 90 ⇒ (n − 1)4 = 80 ⇒ n − 1 = 20 ⇒ n = 21.	[1 Mark] for evaluating n = 21.
Q3	S₁₅ = (15/2)[2(10) + 14(4)] = (15/2)[20 + 56] = (15/2)(76) = 15 × 38 = 570 satellites.	[1 Mark] for formula setup. [1 Mark] for final answer 570.
Q3 (OR)	(n/2)[20 + (n − 1)4] = 960 ⇒ n[2n + 8] = 960 ⇒ 2n² + 8n − 960 = 0 ⇒ n² + 4n − 480 = 0. (n − 20)(n + 24) = 0 ⇒ n = 20 rings (rejecting −24).	[1 Mark] for quadratic equation. [1 Mark] for evaluating n = 20.

Case Study 3 Sports Analytics 4 Marks

Marathon Endurance Training

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An athlete's marathon preparation increases weekly long-run targets progressively:

• Week 1 (a): 5 km baseline
• Weekly increase (d): 2 km/week
• Targets: 5, 7, 9, 11, 13, ...

Q1. Conceptual Understanding 1 Mark

Calculate the athlete's target daily running distance in the 12th week of the cycle.

(A) 25 km

(B) 27 km

(C) 29 km

(D) 31 km

Q2. Application 1 Mark

In which specific week of training will the target distance reach exactly 45 km?

(A) 21st week

(B) 20th week

(C) 22nd week

(D) 23rd week

Q3. Analytical Reasoning 2 Marks

Find the cumulative distance covered on weekly tracking runs during the first 10 weeks.

(A) 120 km

(B) 130 km

(C) 140 km

(D) 150 km

OR (Alternative Q3)

If cumulative distance must reach 437 km, determine the total weeks of training required.

(A) 18 weeks

(B) 19 weeks

(C) 20 weeks

(D) 21 weeks

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁₂ = 5 + 11(2) = 5 + 22 = 27 km.	[1 Mark] for evaluating a₁₂ = 27 km.
Q2	5 + (n − 1)2 = 45 ⇒ (n − 1)2 = 40 ⇒ n − 1 = 20 ⇒ n = 21st week.	[1 Mark] for evaluating n = 21.
Q3	S₁₀ = (10/2)[2(5) + 9(2)] = 5[10 + 18] = 5(28) = 140 km.	[1 Mark] for S_n formula. [1 Mark] for 140 km.
Q3 (OR)	(n/2)[10 + (n − 1)2] = 437 ⇒ n[n + 4] = 437 ⇒ n² + 4n − 437 = 0. (n − 19)(n + 23) = 0 ⇒ n = 19 weeks (rejecting −23).	[1 Mark] for quadratic formulation. [1 Mark] for n = 19 weeks.

Case Study 4 Fintech Savings 4 Marks

Micro-Investment Challenge

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A smart banking app's savings challenge invests weekly sums following an AP:

• Week 1 (a): ₹150 baseline
• Weekly increment (d): ₹50
• Sequence: ₹150, ₹200, ₹250, ₹300, ...

Q1. Conceptual Understanding 1 Mark

Calculate the exact amount of money invested by the algorithm in the 25th week.

(A) ₹1,250

(B) ₹1,300

(C) ₹1,350

(D) ₹1,400

Q2. Application 1 Mark

In which specific week will the weekly investment transfer reach exactly ₹2,000?

(A) 38th week

(B) 36th week

(C) 40th week

(D) 42nd week

Q3. Analytical Reasoning 2 Marks

Find the total accumulated savings invested at the end of the first 20 weeks.

(A) ₹11,500

(B) ₹12,500

(C) ₹13,000

(D) ₹14,200

OR (Alternative Q3)

To accumulate milestone of ₹26,250, how many weeks of this progressive challenge are required?

(A) 25 weeks

(B) 28 weeks

(C) 30 weeks

(D) 32 weeks

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₂₅ = 150 + 24(50) = 150 + 1200 = ₹ 1,350.	[1 Mark] for calculating ₹ 1,350.
Q2	150 + (n − 1)50 = 2000 ⇒ (n − 1)50 = 1850 ⇒ n − 1 = 37 ⇒ n = 38th week.	[1 Mark] for evaluating n = 38.
Q3	S₂₀ = (20/2)[2(150) + 19(50)] = 10[300 + 950] = 10(1250) = ₹ 12,500.	[1 Mark] for S_n setup. [1 Mark] for ₹ 12,500.
Q3 (OR)	(n/2)[300 + (n − 1)50] = 26250 ⇒ 25n² + 125n − 26250 = 0 ⇒ n² + 5n − 1050 = 0. (n − 30)(n + 35) = 0 ⇒ n = 30 weeks (rejecting −35).	[1 Mark] for quadratic formulation. [1 Mark] for solving n = 30 weeks.

Case Study 5 Smart Cities 4 Marks

EV Charging Station Grid

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A smart city administration expands ultra-fast EV charging stations progressively across phases:

• Phase 1 (a): 15 charging stations
• Increase per phase (d): 8 stations
• Phased counts: 15, 23, 31, 39, ...

Q1. Conceptual Understanding 1 Mark

Calculate the exact number of stations that will be newly installed during Phase 15.

(A) 119 stations

(B) 123 stations

(C) 127 stations

(D) 135 stations

Q2. Application 1 Mark

In which specific expansion phase will the administration install exactly 103 charging stations?

(A) Phase 11

(B) Phase 12

(C) Phase 13

(D) Phase 14

Q3. Analytical Reasoning 2 Marks

Find the total cumulative number of charging stations installed across the first 10 phases.

(A) 510 stations

(B) 490 stations

(C) 530 stations

(D) 550 stations

OR (Alternative Q3)

If the developmental target is to establish 938 stations, calculate the number of phases required.

(A) 12 phases

(B) 13 phases

(C) 15 phases

(D) 14 phases

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁₅ = 15 + 14(8) = 15 + 112 = 127 stations.	[1 Mark] for evaluating a₁₅ = 127.
Q2	15 + (n − 1)8 = 103 ⇒ (n − 1)8 = 88 ⇒ n − 1 = 11 ⇒ n = 12.	[1 Mark] for evaluating n = 12.
Q3	S₁₀ = (10/2)[2(15) + 9(8)] = 5[30 + 72] = 5(102) = 510 stations.	[1 Mark] for S₁₀ formula. [1 Mark] for 510 stations.
Q3 (OR)	(n/2)[30 + (n − 1)8] = 938 ⇒ n[4n + 11] = 938 ⇒ 4n² + 11n − 938 = 0. Using quadratic formula: n = 14 phases (rejecting negative root).	[1 Mark] for quadratic formulation. [1 Mark] for solving n = 14.

Case Study 6 Warehouse Logistics 4 Marks

Fulfillment Center Sorting Throughput

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A logistics hub's AI conveyor grid scales daily sorting throughput to handle peak holiday traffic:

• Day 1 (a): 1,200 packages
• Daily increase (d): 150 packages/day
• Progression: 1200, 1350, 1500, 1650, ...

Q1. Conceptual Understanding 1 Mark

Calculate the number of packages sorted by the automated grid on Day 15.

(A) 3,150 packages

(B) 3,300 packages

(C) 3,450 packages

(D) 3,600 packages

Q2. Application 1 Mark

On which operational day will the daily sorting capacity reach exactly 4,200 packages?

(A) Day 21

(B) Day 20

(C) Day 22

(D) Day 24

Q3. Analytical Reasoning 2 Marks

Find the total cumulative number of packages sorted during the first 10 days of the campaign.

(A) 17,500 packages

(B) 18,250 packages

(C) 19,500 packages

(D) 18,750 packages

OR (Alternative Q3)

Determine the exact number of days required to process a cumulative volume of 37,200 packages.

(A) 14 days

(B) 15 days

(C) 16 days

(D) 18 days

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁₅ = 1200 + 14(150) = 1200 + 2100 = 3,300 packages.	[1 Mark] for evaluating a₁₅ = 3,300.
Q2	1200 + (n − 1)150 = 4200 ⇒ (n − 1)150 = 3000 ⇒ n − 1 = 20 ⇒ n = 21.	[1 Mark] for evaluating n = 21.
Q3	S₁₀ = (10/2)[2(1200) + 9(150)] = 5[2400 + 1350] = 5(3750) = 18,750 packages.	[1 Mark] for S₁₀ formula. [1 Mark] for 18,750.
Q3 (OR)	(n/2)[2400 + (n − 1)150] = 37200 ⇒ 75n² + 1125n − 37200 = 0 ⇒ n² + 15n − 496 = 0. (n − 16)(n + 31) = 0 ⇒ n = 16 days (rejecting −31).	[1 Mark] for quadratic equation. [1 Mark] for solving n = 16 days.

Case Study 7 Acoustic Architecture 4 Marks

Amphitheater Seating Rows

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An open-air semicircular amphitheater arranges seating in expanding concentric rows:

• Row 1 (a): 24 seats
• Increase per row (d): 4 seats
• Seating counts: 24, 28, 32, 36, ...

Q1. Conceptual Understanding 1 Mark

Calculate the number of seats installed in the 15th seating row from the stage.

(A) 76 seats

(B) 80 seats

(C) 84 seats

(D) 88 seats

Q2. Application 1 Mark

If the outermost row contains 120 seats, find the total number of concentric rows.

(A) 22 rows

(B) 23 rows

(C) 24 rows

(D) 25 rows

Q3. Analytical Reasoning 2 Marks

Calculate the total seating capacity of the amphitheater if built with exactly 20 rows.

(A) 1,240 seats

(B) 1,200 seats

(C) 1,280 seats

(D) 1,320 seats

OR (Alternative Q3)

To accommodate a booking requiring a minimum of 952 seats, calculate the minimum number of rows.

(A) 16 rows

(B) 17 rows

(C) 18 rows

(D) 19 rows

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁₅ = 24 + 14(4) = 24 + 56 = 80 seats.	[1 Mark] for a₁₅ = 80.
Q2	24 + (n − 1)4 = 120 ⇒ (n − 1)4 = 96 ⇒ n − 1 = 24 ⇒ n = 25 rows.	[1 Mark] for n = 25.
Q3	S₂₀ = (20/2)[2(24) + 19(4)] = 10[48 + 76] = 10(124) = 1,240 seats.	[1 Mark] for formula setup. [1 Mark] for 1,240 seats.
Q3 (OR)	(n/2)[48 + (n − 1)4] = 952 ⇒ 2n² + 22n − 952 = 0 ⇒ n² + 11n − 476 = 0. (n − 17)(n + 28) = 0 ⇒ n = 17 rows (rejecting −28).	[1 Mark] for quadratic formulation. [1 Mark] for n = 17 rows.

Case Study 8 Hydrology & Climate 4 Marks

Groundwater Depletion Monitoring

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Hydrogeological scientists record groundwater depletion rates in an agrarian district:

• Year 1 depth (a): 40 meters
• Annual drop rate (d): 1.5 meters/year
• Depletion depths: 40, 41.5, 43, 44.5, ...

Q1. Conceptual Understanding 1 Mark

Calculate the estimated depth of the water table level below the surface in the 11th year.

(A) 53.5 meters

(B) 55.0 meters

(C) 56.5 meters

(D) 57.0 meters

Q2. Application 1 Mark

In which specific survey year will the critical 67-meter dry-well boundary be breached?

(A) Year 19

(B) Year 18

(C) Year 20

(D) Year 21

Q3. Analytical Reasoning 2 Marks

Find the sum of the water table depths recorded across the first 12 years of the study.

(A) 565 meters

(B) 572 meters

(C) 579 meters

(D) 584 meters

OR (Alternative Q3)

If sum of recorded depths over consecutive years starting from Year 1 equals 414 m, find the total years included.

(A) 8 years

(B) 10 years

(C) 11 years

(D) 9 years

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	a₁₁ = 40 + 10(1.5) = 40 + 15 = 55 meters.	[1 Mark] for a₁₁ = 55 m.
Q2	40 + (n − 1)1.5 = 67 ⇒ (n − 1)1.5 = 27 ⇒ n − 1 = 18 ⇒ n = 19th year.	[1 Mark] for n = 19.
Q3	S₁₂ = (12/2)[2(40) + 11(1.5)] = 6[80 + 16.5] = 6(96.5) = 579 meters.	[1 Mark] for S₁₂ formula. [1 Mark] for 579 meters.
Q3 (OR)	(n/2)[80 + (n − 1)1.5] = 414 ⇒ 1.5n² + 78.5n − 828 = 0 ⇒ 3n² + 157n − 1656 = 0. Solving gives n = 9 years (rejecting negative root).	[1 Mark] for quadratic formulation. [1 Mark] for solving n = 9 years.