Arithmetic Progressions: Case Study PYQs with Answers

Yes; d = 6; next term = 30.

12 − 6 = 18 − 12 = 24 − 18 = 6.

Equal consecutive differences show that the pattern forms an AP.

d = 6.

Next term = 24 + 6 = 30.

aₙ = 6n; 60 stitches in the 10th row.

Here, a = 6 and d = 6.

aₙ = a + (n − 1)d.

aₙ = 6 + (n − 1)6 = 6n.

a₁₀ = 6 × 10 = 60 stitches.

4, 6, 8, …; aₙ = 2n + 2.

The figures contain 4, 6 and 8 triangles.

a = 4 and d = 2.

aₙ = a + (n − 1)d.

aₙ = 4 + 2(n − 1) = 2n + 2.

Figure 8.

Matchstick counts are 12, 19, 26, …

a = 12 and d = 7.

61 = 12 + 7(n − 1).

7(n − 1) = 49.

n = 8.

₹14,000.

Here, a = 2000, d = 500 and n = 25.

aₙ = a + (n − 1)d.

aₙ = 2000 + (25 − 1) × (500).

= 14000 rupees.

₹2,00,000.

Here, a = 2000, d = 500 and n = 25.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{25}\text{)/(}}{\text{2}\text{)}}$[2 × 2000 + (25 − 1) × (500)].

= 200000 rupees.

a + 3d = 80,000.

aₙ = a + (n − 1)d.

a₄ = a + 3d = 80,000.

a + 6d = 1,10,000.

a₇ = a + (7 − 1)d.

a + 6d = 1,10,000.

50,000 cars.

Let a be first-year production and d the annual increase.

a + 3d = 80000.

a + 6d = 110000.

Subtracting, 3d = 30000.

d = 10000.

a = 80000 − 3 × 10000 = 50000.

10,000 cars per year.

a + 3d = 80,000.

a + 6d = 1,10,000.

Subtracting, 3d = 30,000.

d = 10,000.

₹3,800.

Here, a = 2000, d = 200 and n = 10.

aₙ = a + (n − 1)d.

aₙ = 2000 + (10 − 1) × (200).

= 3800 rupees.

₹29,000.

Here, a = 2000, d = 200 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 2000 + (10 − 1) × (200)].

= 29000 rupees.

50 instalments.

Here, a = 2,000, d = 200 and Sₙ = 3,45,000.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

3,45,000 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[4,000 + 200(n − 1)].

n² + 19n − 3,450 = 0.

(n − 50)(n + 69) = 0.

n = 50; reject −69 because a count must be positive.

₹2,88,000.

Here, a = 2000, d = 200 and n = 45.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{45}\text{)/(}}{\text{2}\text{)}}$[2 × 2000 + (45 − 1) × (200)].

= 288000 rupees.

1, 3, 5, 7, …

Count shaded squares row-wise: 1, 3, 5, 7, …

a = 1 and d = 2.

2, 6, 10, 14, …

Count unshaded triangles row-wise: 2, 6, 10, 14, …

a = 2 and d = 4.

900 cm².

For squares, a = 1 and d = 2.

S₁₅ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$[2 × 1 + 14 × 2] = 225 squares.

Area of one square = 2² = 4 cm².

Shaded area = 225 × 4 = 900 cm².

Sₙ = 2n²; S₁₀ = 200 triangles.

For triangles, a = 2 and d = 4.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[4 + 4(n − 1)] = 2n².

S₁₀ = 2 × 10² = 200 triangles.

Yes; a = ₹240 and d = ₹60.

300 − 240 = 360 − 300 = 60.

Equal consecutive differences show an AP.

a = ₹240 and d = ₹60.

The 8th month.

660 = 240 + 60(n − 1).

n − 1 = 7.

n = 8.

₹1,080.

Here, a = 240, d = 60 and n = 15.

aₙ = a + (n − 1)d.

aₙ = 240 + (15 − 1) × (60).

= 1080 rupees.

₹5,100.

Here, a = 240, d = 60 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 240 + (10 − 1) × (60)].

= 5100 rupees.

a = 1 and d = 4.

Square counts are 1, 5, 9, …

a = 1 and d = 5 − 1 = 4.

a = 4 and d = 12.

Stick counts are 4, 16, 28, …

a = 4 and d = 16 − 4 = 12.

37 squares and 112 matchsticks.

aₙ = a + (n − 1)d.

Squares = 1 + 9 × 4 = 37.

Sticks = 4 + 9 × 12 = 112.

m = 8; 29 squares.

88 = 4 + (m − 1)12.

m = 8.

Squares = 1 + 7 × 4 = 29.

24.

Given aₙ = 20 + 4n.

a₁ = 20 + 4 × 1 = 24.

The 23rd spot.

112 = 20 + 4n.

4n = 92.

n = 23.

420.

a = 24 and d = 4.

S₁₀ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 24 + 9 × 4].

= 5 × 84 = 420.

4n + 12 (n ≥ 3).

Replace the term index by n − 2.

aₙ₋₂ = 20 + 4(n − 2).

aₙ₋₂ = 4n + 12.

A spot index is positive, so n ≥ 3.

250 road rollers.

Let a be first-year production and d the annual increase.

a + 5d = 800.

a + 8d = 1130.

Subtracting, 3d = 330.

d = 110.

a = 800 − 5 × 110 = 250.

110 road rollers per year.

a + 5d = 800.

a + 8d = 1,130.

3d = 330.

d = 110.

1,020 road rollers.

Here, a = 250, d = 110 and n = 8.

aₙ = a + (n − 1)d.

aₙ = 250 + (8 − 1) × (110).

= 1020 road rollers.

3,150 road rollers.

Here, a = 250, d = 110 and n = 6.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{6}\text{)/(}}{\text{2}\text{)}}$[2 × 250 + (6 − 1) × (110)].

= 3150 road rollers.

17 km each day.

Here, a = 3, d = 2 and n = 8.

aₙ = a + (n − 1)d.

aₙ = 3 + (8 − 1) × (2).

= 17 km.

The 22nd week.

45 = 3 + 2(n − 1).

n − 1 = 21.

n = 22.

715 km.

The daily distances across weeks have a = 3 and d = 2.

S₁₁ = $\frac{\text{(}\text{11}\text{)/(}}{\text{2}\text{)}}$[2 × 3 + 10 × 2] = 143 km.

This counts one practice day from each week.

Total for five days each week = 5 × 143 = 715 km.

14 weeks.

Now a = 3 and d = 3.

42 = 3 + (n − 1)3.

n − 1 = 13.

n = 14.

51, 49, 47, …, 31 seconds.

The starting time is 51 seconds.

Subtract 2 seconds for each successive day.

The AP is 51, 49, 47, …, 31.

Day 11 under the official scheme’s indexing.

a = 51 and d = −2.

31 = 51 + (n − 1)(−2).

n − 1 = 10.

n = 11.

The scheme counts 51 seconds as Day 1: Day 11 follows 10 daily improvements.

aₙ = 53 − 2n.

aₙ = a + (n − 1)d.

aₙ = 51 − 2(n − 1).

aₙ = 53 − 2n.

Day 16 under the same indexing.

21 = 51 − 2(n − 1).

2(n − 1) = 30.

n = 16.

The 16th listed day follows 15 daily improvements.

₹55,000.

Here, a = 10000, d = 5000 and n = 10.

aₙ = a + (n − 1)d.

aₙ = 10000 + (10 − 1) × (5000).

= 55000 rupees.

The 7th instalment.

40,000 = 10,000 + (n − 1)5,000.

n − 1 = 6.

n = 7.

20 instalments.

11,50,000 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[20,000 + (n − 1)5,000].

n² + 3n − 460 = 0.

(n − 20)(n + 23) = 0.

n = 20; reject −23.

The 10th instalment.

3,25,000 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[20,000 + (n − 1)5,000].

n² + 3n − 130 = 0.

(n − 10)(n + 13) = 0.

n = 10; reject −13.

44 cm, 54 cm, 64 cm, 74 cm, 84 cm.

h₁ = 44 cm.

Add 10 cm for each subsequent seat.

h₂ = 54 cm, h₃ = 64 cm, h₄ = 74 cm, h₅ = 84 cm.

Yes; a = 44 cm and d = 10 cm.

54 − 44 = 64 − 54 = 74 − 64 = 84 − 74 = 10.

Equal consecutive differences show an AP.

a = 44 cm and d = 10 cm.

12 chairs in the question’s seat-height model.

hₙ = 44 + (n − 1)10.

44 + (n − 1)10 ≤ 160.

n ≤ 12.6.

The greatest whole-number count is 12.

h₁₂ = 154 cm ≤ 160 cm; h₁₃ = 164 cm > 160 cm.

The marking scheme models stack height using the labelled seat heights.

No; h₁₅ = 184 cm > 180 cm.

h₁₅ = 44 + (15 − 1)10.

h₁₅ = 184 cm.

184 cm exceeds the limit of 180 cm.

450 m, 500 m and 550 m.

Use the stated sequence 300, 350, 400, … metres.

a = 300 and d = 50.

a₄ = 300 + 3 × 50 = 450 m.

a₅ = 300 + 4 × 50 = 500 m.

a₆ = 300 + 5 × 50 = 550 m.

650 m.

Here, a = 300, d = 50 and n = 8.

aₙ = a + (n − 1)d.

aₙ = 300 + (8 − 1) × (50).

= 650 m.

5,250 m.

Here, a = 300, d = 50 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 300 + (10 − 1) × (50)].

= 5250 m.

2,550 m (corrected arithmetic; scheme prints 2,250 m).

Here, a = 300, d = 50 and n = 6.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{6}\text{)/(}}{\text{2}\text{)}}$[2 × 300 + (6 − 1) × (50)].

= 2550 m.

Source correction: CBSE 2025 Standard papers 30/1/1 (Q36), 30/1/2 (Q38) and 30/1/3 (Q37) have schemes that print 2,250 m after the same expression; the calculation gives 2,550 m.

The representative scheme file is X_MS_041_Mathematics Standard_30-1-1_2024-25.pdf. All three filenames are recorded under Case 15 sources.

1,550 m.

Here, a = 200, d = 150 and n = 10.

aₙ = a + (n − 1)d.

aₙ = 200 + (10 − 1) × (150).

= 1550 m.

1,500 m.

a₂₅ − a₁₅ = (a + 24d) − (a + 14d).

= 10d.

= 10 × 150 = 1,500 m.

480 seconds (8 minutes).

For this part, count poles from the top, as the question states.

The nearest pole is 300 m from the top; consecutive gaps are 150 m.

Distance to 15th pole = 300 + 14 × 150 = 2,400 m.

Time = distance ÷ speed.

Time = $\frac{\text{(}\text{2,400}\text{)/(}}{\text{5}\text{)}}$ = 480 seconds = 8 minutes.

31 poles.

Last pole is 5,000 − 300 = 4,700 m from the base.

4,700 = 200 + (n − 1)150.

n − 1 = 30.

n = 31.

19 triangles.

There are 10 rows from the top.

a = 1 and d = 2.

a₁₀ = 1 + 9 × 2 = 19 triangles.

13 triangles.

Fourth row from bottom = row 10 − 4 + 1 = 7 from top.

a₇ = 1 + 6 × 2 = 13 triangles.

64 triangles.

Here, a = 1, d = 2 and n = 8.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{8}\text{)/(}}{\text{2}\text{)}}$[2 × 1 + (8 − 1) × (2)].

= 64 triangles.

68 more triangles.

For 1, 3, 5, …, Sₙ = n².

S₁₀ = 100 and S₄ = 16.

Rows 5 to 10 contain 100 − 16 = 84 triangles.

Required difference = 84 − 16 = 68 triangles.

438 m.

Here, a = 400, d = 7.6 and n = 6.

aₙ = a + (n − 1)d.

aₙ = 400 + (6 − 1) × (7.6).

= 438 m.

30.4 m longer.

a₈ − a₄ = (a + 7d) − (a + 3d).

= 4d.

= 4 × 7.6 = 30.4 m.

2,514 m.

Here, a = 400, d = 7.6 and n = 6.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{6}\text{)/(}}{\text{2}\text{)}}$[2 × 400 + (6 − 1) × (7.6)].

= 2514 m.

2,190 m.

a₄ = 400 + 3 × 7.6 = 422.8 m.

a₈ = 400 + 7 × 7.6 = 453.2 m.

Number of included lanes = 8 − 4 + 1 = 5.

Sum = $\frac{\text{(}\text{5}\text{)/(}}{\text{2}\text{)}}$(422.8 + 453.2) = 2,190 m.

49 houses.

Count = 49 − 1 + 1 = 49 houses.

a = 1 and d = 1.

The AP is 1, 2, 3, …, 49.

a = 1 and d = 2 − 1 = 1.

1,225.

Here, a = 1, l = 49 and n = 49.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$(a + l).

S₄₉ = $\frac{\text{(}\text{49}\text{)/(}}{\text{2}\text{)}}$(1 + 49) = 1,225.

315.

The scheme interprets “between” strictly: 16, 17, …, 29.

Number of terms = 29 − 16 + 1 = 14.

Sum = $\frac{\text{(}\text{14}\text{)/(}}{\text{2}\text{)}}$(16 + 29) = 315.

a = 72 and d = −4.

a = 72.

d = 68 − 72 = −4.

The 5th coach.

56 = 72 + (n − 1)(−4).

n − 1 = 4.

n = 5.

540 passengers.

Here, a = 72, d = -4 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 72 + (10 − 1) × (-4)].

= 540 passengers.

36 passengers.

Here, a = 72, d = -4 and n = 10.

aₙ = a + (n − 1)d.

aₙ = 72 + (10 − 1) × (-4).

= 36 passengers.

182.

Smallest three-digit multiple of 13 = 104.

The AP is 104, 117, 130, …

a₇ = 104 + 6 × 13 = 182.

65.

a₁₄ − a₉ = (a + 13d) − (a + 8d).

= 5 × 13 = 65.

884.

Largest three-digit multiple of 13 = 988.

Counting backwards, d = −13.

9th value from end = 988 − 8 × 13 = 884.

37,674.

a = 104, d = 13 and l = 988.

988 = 104 + (n − 1)13.

n = 69.

S₆₉ = $\frac{\text{(}\text{69}\text{)/(}}{\text{2}\text{)}}$(104 + 988) = 37,674.

₹3,900.

Here, a = 1000, d = 100 and n = 30.

aₙ = a + (n − 1)d.

aₙ = 1000 + (30 − 1) × (100).

= 3900 rupees.

₹4,900.

Here, a = 1000, d = 100 and n = 40.

aₙ = a + (n − 1)d.

aₙ = 1000 + (40 − 1) × (100).

= 4900 rupees.

₹44,500.

S₃₀ = $\frac{\text{(}\text{30}\text{)/(}}{\text{2}\text{)}}$[2 × 1,000 + 29 × 100].

S₃₀ = ₹73,500.

Remaining = 1,18,000 − 73,500 = ₹44,500.

19 : 49.

a₁₀ = 1,000 + 9 × 100 = ₹1,900.

a₄₀ = 1,000 + 39 × 100 = ₹4,900.

Ratio = 1,900 : 4,900 = 19 : 49.

₹5,000, ₹5,500 and ₹6,000.

Start with ₹5,000.

Add ₹500 for each next instalment.

The first three are ₹5,000, ₹5,500 and ₹6,000.

₹10,000.

Here, a = 5000, d = 500 and n = 11.

aₙ = a + (n − 1)d.

aₙ = 5000 + (11 − 1) × (500).

= 10000 rupees.

40 instalments.

5,90,000 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[10,000 + (n − 1)500].

n² + 19n − 2,360 = 0.

(n − 40)(n + 59) = 0.

n = 40; reject −59.

₹2,02,500.

S₃₁ = $\frac{\text{(}\text{31}\text{)/(}}{\text{2}\text{)}}$[10,000 + 30 × 500].

S₃₁ = ₹3,87,500.

Remaining = 5,90,000 − 3,87,500 = ₹2,02,500.

60, 57, 54, 51, 48 seconds; d = −3.

The first five times are 60, 57, 54, 51 and 48 seconds.

57 − 60 = 54 − 57 = 51 − 54 = 48 − 51 = −3.

Equal consecutive differences show an AP.

45 seconds.

Here, a = 60, d = -3 and n = 6.

aₙ = a + (n − 1)d.

aₙ = 60 + (6 − 1) × (-3).

= 45 seconds.

Day 11.

Day 1 is explicitly given as 60 seconds.

30 = 60 + (n − 1)(−3).

n − 1 = 10.

n = 11.

Day 11.

Now a = 60 and d = −3.2.

28 = 60 + (n − 1)(−3.2).

n − 1 = $\frac{\text{(}\text{32}\text{)/(}}{\text{3.2}\text{)}}$ = 10.

n = 11.

Yes; a = 4 and d = 4.

8 − 4 = 12 − 8 = 4.

Equal consecutive differences show an AP.

a = 4 and d = 4.

aₙ = 4n.

aₙ = a + (n − 1)d.

aₙ = 4 + (n − 1)4 = 4n.

10 squares.

220 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2 × 4 + (n − 1)4].

n(n + 1) = 110.

(n − 10)(n + 11) = 0.

n = 10; reject −11.

No.

Sₙ = 2n(n + 1).

100 = 2n(n + 1).

n² + n − 50 = 0.

n = $\frac{\text{(}\text{−1 ± √201}\text{)/(}}{\text{2}\text{)}}$.

Neither root is a positive integer, so exactly 100 dots cannot complete this pattern.

40, 46, 52, 58, … cm.

The first step is at 40 cm.

The folded spacing is 6 cm.

The AP is 40, 46, 52, 58, … cm.

40, 70, 100, 130, … cm.

The first step remains at 40 cm.

The extended spacing is 30 cm.

The AP is 40, 70, 100, 130, … cm.

Position I: 94 cm; Position II: 310 cm.

Ten steps give nine gaps above the first step.

Position I = 40 + 9 × 6 = 94 cm.

Position II = 40 + 9 × 30 = 310 cm.

238 cm.

Six gaps are extended to 30 cm.

Height to 7th step = 40 + 6 × 30 = 220 cm.

The remaining three gaps stay folded at 6 cm.

Total height = 220 + 3 × 6 = 238 cm.

Yes; a = ₹200 and d = ₹50.

250 − 200 = 300 − 250 = 50.

Equal consecutive differences show an AP.

a = ₹200 and d = ₹50.

The 17th month.

1,000 = 200 + (n − 1)50.

n − 1 = 16.

n = 17.

₹13,500.

Here, a = 200, d = 50 and n = 20.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{20}\text{)/(}}{\text{2}\text{)}}$[2 × 200 + (20 − 1) × (50)].

= 13500 rupees.

26 months.

21,450 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[400 + (n − 1)50].

n² + 7n − 858 = 0.

(n − 26)(n + 33) = 0.

n = 26; reject −33.

150 road rollers.

Let a be first-year production and d the annual increase.

a + 3d = 270.

a + 9d = 510.

Subtracting, 6d = 240.

d = 40.

a = 270 − 3 × 40 = 150.

430 road rollers.

Here, a = 150, d = 40 and n = 8.

aₙ = a + (n − 1)d.

aₙ = 150 + (8 − 1) × (40).

= 430 road rollers.

1,500 road rollers.

Here, a = 150, d = 40 and n = 6.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{6}\text{)/(}}{\text{2}\text{)}}$[2 × 150 + (6 − 1) × (40)].

= 1500 road rollers.

19 : 43.

a₂ = 150 + 40 = 190.

a₈ = 150 + 7 × 40 = 430.

Ratio = 190 : 430 = 19 : 43.

1,200 cars.

Let a be first-year production and d the annual increase.

a + 3d = 1800.

a + 7d = 2600.

Subtracting, 4d = 800.

d = 200.

a = 1800 − 3 × 200 = 1200.

3,400 cars.

Here, a = 1200, d = 200 and n = 12.

aₙ = a + (n − 1)d.

aₙ = 1200 + (12 − 1) × (200).

= 3400 cars.

21,000 cars.

Here, a = 1200, d = 200 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 1200 + (10 − 1) × (200)].

= 21000 cars.

13 years.

31,200 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2 × 1,200 + (n − 1)200].

n² + 11n − 312 = 0.

(n − 13)(n + 24) = 0.

n = 13; reject −24.

120 seats.

Here, a = 30, d = 10 and n = 10.

aₙ = a + (n − 1)d.

aₙ = 30 + (10 − 1) × (10).

= 120 seats.

15 rows.

a = 30 and d = 10.

1,500 = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[60 + (n − 1)10].

n² + 5n − 300 = 0.

(n − 15)(n + 20) = 0.

n = 15; reject −20.

750 seats.

S₁₀ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × 30 + 9 × 10] = 750 seats.

Seats left = 1,500 − 750 = 750.

110 seats.

Middle row = $\frac{\text{(}\text{17 + 1}\text{)/(}}{\text{2}\text{)}}$ = 9th row.

a₉ = 30 + 8 × 10 = 110 seats.

The 12th year.

Let a be first-year production and d the annual increase.

a + 5d = 16000.

a + 8d = 22600.

Subtracting, 3d = 6600.

d = 2200.

a = 16000 − 5 × 2200 = 5000.

29,200 = 5,000 + (n − 1)2,200.

n = 12.

20,400 TV sets.

Here, a = 5000, d = 2200 and n = 8.

aₙ = a + (n − 1)d.

aₙ = 5000 + (8 − 1) × (2200).

= 20400 TV sets.

21,600 TV sets.

Here, a = 5000, d = 2200 and n = 3.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{3}\text{)/(}}{\text{2}\text{)}}$[2 × 5000 + (3 − 1) × (2200)].

= 21600 TV sets.

6,600 TV sets.

a₇ − a₄ = (a + 6d) − (a + 3d).

= 3d = 3 × 2,200 = 6,600 TV sets.

160 throws.

Here, a = 40, d = 12 and n = 11.

aₙ = a + (n − 1)d.

aₙ = 40 + (11 − 1) × (12).

= 160 throws.

8.10 m.

Initial distance = 7.56 m.

Weekly increase = 9 cm = 0.09 m.

Six completed weeks give six increases.

Distance = 7.56 + 6 × 0.09 = 8.10 m.

After 40 weekly improvements; labelled the 41st week in the official scheme.

Let k count completed weekly improvements.

11.16 = 7.56 + 0.09k.

k = $\frac{\text{(}\text{11.16 − 7.56}\text{)/(}}{\text{0.09}\text{)}}$ = 40.

If the initial throw is term 1, this is term 41.

The scheme reports 41 weeks by term numbering; 40 weeks have elapsed since the initial throw.

1,860 throws.

Here, a = 40, d = 12 and n = 15.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{15}\text{)/(}}{\text{2}\text{)}}$[2 × 40 + (15 − 1) × (12)].

= 1860 throws.

3, 6, 9, …, 24; d = 3.

Eight layer counts: 3, 6, 9, 12, 15, 18, 21, 24.

d = 6 − 3 = 3.

No.

34 = 3 + (n − 1)3.

n = $\frac{\text{(}\text{34}\text{)/(}}{\text{3}\text{)}}$, which is not a positive integer.

No complete layer has exactly 34 jars.

Sₙ = $\frac{\text{(}\text{3n(n + 1)}\text{)/(}}{\text{2}\text{)}}$; S₈ = 108 jars.

Following the marking scheme, n counts the successive layers.

a = 3 and d = 3.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[6 + (n − 1)3].

Sₙ = $\frac{\text{(}\text{3n(n + 1)}\text{)/(}}{\text{2}\text{)}}$.

S₈ = $\frac{\text{(}\text{3 × 8 × 9}\text{)/(}}{\text{2}\text{)}}$ = 108 jars.

18 jars.

The new AP is 6, 9, 12, …

a = 6 and d = 3.

a₅ = 6 + (5 − 1)3 = 18 jars.

0.

Aryan: d₁ = −2 − (−5) = 3.

Roshan: d₂ = 184 − 187 = −3.

Sum = 3 + (−3) = 0.

88.

Here, a = 187, d = -3 and n = 34.

aₙ = a + (n − 1)d.

aₙ = 187 + (34 − 1) × (-3).

= 88.

85.

Here, a = -5, d = 3 and n = 10.

Sₙ = $\frac{\text{(}\text{n}\text{)/(}}{\text{2}\text{)}}$[2a + (n − 1)d].

Sₙ = $\frac{\text{(}\text{10}\text{)/(}}{\text{2}\text{)}}$[2 × -5 + (10 − 1) × (3)].

= 85.

The 33rd term of each; common value = 91.

Equate terms at the same position n.

−5 + 3(n − 1) = 187 − 3(n − 1).

6(n − 1) = 192.

n = 33.

Common value = −5 + 32 × 3 = 91.

4 trees.

One section of Class n plants 2n + 3 trees.

Two sections plant 2(2n + 3) = 4n + 6 trees.

Class totals are 10, 14, 18, …

d = 14 − 10 = 4 trees.

30 trees.

Each section plants 2 × 6 + 3 = 15 trees.

Two sections plant 2 × 15 = 30 trees.

384 trees.

Use class totals: a = 10, d = 4 and n = 12.

S₁₂ = $\frac{\text{(}\text{12}\text{)/(}}{\text{2}\text{)}}$[2 × 10 + 11 × 4].

= 6 × 64 = 384 trees.

Class VII.

34 = 10 + (n − 1)4.

n − 1 = 6.

n = 7, so the class is VII.