Arithmetic Progressions - Complete Solutions

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Key Formulas

Exercise 5.1 Solutions

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.

Yes, this forms an AP. The fares would be: 15, 23, 31, 39, ... with common difference d = 8.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

No, this does not form an AP. If we start with V amount of air, after first removal: 3V/4, after second: (3/4)²V, after third: (3/4)³V, etc. This is a geometric progression, not arithmetic.

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

Yes, this forms an AP. The costs would be: 150, 200, 250, 300, ... with common difference d = 50.

(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.

No, this does not form an AP. With compound interest, the amounts would be: 10000, 10800, 11664, 12597.12, ... which is a geometric progression, not arithmetic.

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

First four terms: 10, 20, 30, 40

(ii) a = -2, d = 0

First four terms: -2, -2, -2, -2

(iii) a = 4, d = -3

First four terms: 4, 1, -2, -5

(iv) a = -1, d = 1/2

First four terms: -1, -1/2, 0, 1/2

(v) a = -1.25, d = -0.25

First four terms: -1.25, -1.50, -1.75, -2.00

3. For the following APs, write the first term and the common difference:

(i) 3, 1, -1, -3, ...

First term a = 3, Common difference d = -2

(ii) -5, -1, 3, 7, ...

First term a = -5, Common difference d = 4

(iii) 1/3, 5/3, 9/3, 13/3, ...

First term a = 1/3, Common difference d = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, ...

First term a = 0.6, Common difference d = 1.1

4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, ...

Not an AP (4-2 = 2, 8-4 = 4, differences are not equal)

(ii) 2, 5/2, 3, 7/2, ...

Yes, it's an AP with d = 1/2. Next three terms: 4, 9/2, 5

(iii) -1.2, -3.2, -5.2, -7.2, ...

Yes, it's an AP with d = -2. Next three terms: -9.2, -11.2, -13.2

(iv) -10, -6, -2, 2, ...

Yes, it's an AP with d = 4. Next three terms: 6, 10, 14

(v) 3, 3+√2, 3+2√2, 3+3√2, ...

Yes, it's an AP with d = √2. Next three terms: 3+4√2, 3+5√2, 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222, ...

Not an AP (0.22-0.2 = 0.02, 0.222-0.22 = 0.002, differences are not equal)

(vii) 0, -4, -8, -12, ...

Yes, it's an AP with d = -4. Next three terms: -16, -20, -24

(viii) -1/2, -1/2, -1/2, -1/2, ...

Yes, it's an AP with d = 0. Next three terms: -1/2, -1/2, -1/2

(ix) 1, 3, 9, 27, ...

Not an AP (3-1 = 2, 9-3 = 6, differences are not equal)

(x) a, 2a, 3a, 4a, ...

Yes, it's an AP with d = a. Next three terms: 5a, 6a, 7a

(xi) a, a², a³, a⁴, ...

Not an AP (unless a = 1)

(xii) √2, √8, √18, √32, ...

Yes, it's an AP with d = √2. Next three terms: √50, √72, √98

(xiii) √3, √6, √9, √12, ...

Not an AP (√6-√3 ≠ √9-√6)

(xiv) 1², 3², 5², 7², ...

Not an AP (9-1 = 8, 25-9 = 16, differences are not equal)

(xv) 1², 5², 7², 73, ...

Not an AP (25-1 = 24, 49-25 = 24, 73-49 = 24, but 73 is not 9²)

Exercise 5.2 Solutions

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and aₙ the nth term of the AP:

	a	d	n	aₙ
(i)	7	3	8	28
(ii)	-18	2	10	0
(iii)	46	-3	18	-5
(iv)	-18.9	2.5	10	3.6
(v)	3.5	0	105	3.5

2. Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, ..., is

a = 10, d = -3, n = 30

a₃₀ = 10 + (30-1)(-3) = 10 - 87 = -77

Correct choice: (C) -77

(ii) 11th term of the AP: -3, -1/2, 2, ..., is

a = -3, d = 2.5, n = 11

a₁₁ = -3 + (11-1)(2.5) = -3 + 25 = 22

Correct choice: (B) 22

3. In the following APs, find the missing terms in the boxes:

(i) 2, ▢, 26

Let the missing term be x. Then x - 2 = 26 - x ⇒ 2x = 28 ⇒ x = 14

Answer: 2, 14, 26

(ii) ▢, 13, ▢, 3

Let the AP be: a, 13, b, 3

13 - a = b - 13 = 3 - b

From b - 13 = 3 - b ⇒ 2b = 16 ⇒ b = 8

From 13 - a = 8 - 13 ⇒ 13 - a = -5 ⇒ a = 18

Answer: 18, 13, 8, 3

(iii) 5, ▢, ▢, 9½

a = 5, a₄ = 9.5, n = 4

9.5 = 5 + 3d ⇒ 3d = 4.5 ⇒ d = 1.5

a₂ = 5 + 1.5 = 6.5, a₃ = 6.5 + 1.5 = 8

Answer: 5, 6.5, 8, 9.5

(iv) -4, ▢, ▢, ▢, 6

a = -4, a₅ = 6, n = 5

6 = -4 + 4d ⇒ 4d = 10 ⇒ d = 2.5

a₂ = -4 + 2.5 = -1.5, a₃ = -1.5 + 2.5 = 1, a₄ = 1 + 2.5 = 3.5

Answer: -4, -1.5, 1, 3.5, 6

(v) ▢, 38, ▢, ▢, ▢, -22

a₂ = 38, a₆ = -22

a₂ = a + d = 38, a₆ = a + 5d = -22

Subtracting: 4d = -60 ⇒ d = -15

a = 38 - (-15) = 53

a₃ = 38 - 15 = 23, a₄ = 23 - 15 = 8, a₅ = 8 - 15 = -7

Answer: 53, 38, 23, 8, -7, -22

4. Which term of the AP: 3, 8, 13, 18, ..., is 78?

a = 3, d = 5, aₙ = 78

78 = 3 + (n-1)5 ⇒ 75 = (n-1)5 ⇒ n-1 = 15 ⇒ n = 16

78 is the 16th term.

5. Find the number of terms in each of the following APs:

(i) 7, 13, 19, ..., 205

a = 7, d = 6, aₙ = 205

205 = 7 + (n-1)6 ⇒ 198 = (n-1)6 ⇒ n-1 = 33 ⇒ n = 34

There are 34 terms.

(ii) 18, 15½, 13, ..., -47

a = 18, d = -2.5, aₙ = -47

-47 = 18 + (n-1)(-2.5) ⇒ -65 = (n-1)(-2.5) ⇒ n-1 = 26 ⇒ n = 27

There are 27 terms.

6. Check whether -150 is a term of the AP: 11, 8, 5, 2, ...

a = 11, d = -3

Let -150 = 11 + (n-1)(-3) ⇒ -161 = (n-1)(-3) ⇒ n-1 = 161/3 ≈ 53.67

Since n must be a positive integer, -150 is not a term of this AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

a₁₁ = a + 10d = 38

a₁₆ = a + 15d = 73

Subtracting: 5d = 35 ⇒ d = 7

a = 38 - 10(7) = -32

a₃₁ = -32 + 30(7) = -32 + 210 = 178

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

a₃ = a + 2d = 12

a₅₀ = a + 49d = 106

Subtracting: 47d = 94 ⇒ d = 2

a = 12 - 2(2) = 8

a₂₉ = 8 + 28(2) = 8 + 56 = 64

9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

a₃ = a + 2d = 4

a₉ = a + 8d = -8

Subtracting: 6d = -12 ⇒ d = -2

a = 4 - 2(-2) = 8

Let aₙ = 0 ⇒ 8 + (n-1)(-2) = 0 ⇒ 8 - 2n + 2 = 0 ⇒ 10 = 2n ⇒ n = 5

The 5th term is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

a₁₇ - a₁₀ = 7

(a + 16d) - (a + 9d) = 7 ⇒ 7d = 7 ⇒ d = 1

The common difference is 1.

11. Which term of the AP: 3, 15, 27, 39, ..., will be 132 more than its 54th term?

a = 3, d = 12

a₅₄ = 3 + 53(12) = 3 + 636 = 639

We need term = 639 + 132 = 771

771 = 3 + (n-1)12 ⇒ 768 = (n-1)12 ⇒ n-1 = 64 ⇒ n = 65

The 65th term is 132 more than the 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Let the first AP have first term a and common difference d.

Let the second AP have first term A and common difference d.

100th term difference: (a + 99d) - (A + 99d) = a - A = 100

1000th term difference: (a + 999d) - (A + 999d) = a - A = 100

The difference between their 1000th terms is also 100.

13. How many three-digit numbers are divisible by 7?

Smallest 3-digit number divisible by 7: 105

Largest 3-digit number divisible by 7: 994

This forms an AP: 105, 112, 119, ..., 994

a = 105, d = 7, aₙ = 994

994 = 105 + (n-1)7 ⇒ 889 = (n-1)7 ⇒ n-1 = 127 ⇒ n = 128

There are 128 three-digit numbers divisible by 7.

14. How many multiples of 4 lie between 10 and 250?

First multiple of 4 after 10: 12

Last multiple of 4 before 250: 248

This forms an AP: 12, 16, 20, ..., 248

a = 12, d = 4, aₙ = 248

248 = 12 + (n-1)4 ⇒ 236 = (n-1)4 ⇒ n-1 = 59 ⇒ n = 60

There are 60 multiples of 4 between 10 and 250.

15. For what value of n, are the nth terms of two APs: 63, 65, 67, ..., and 3, 10, 17, ..., equal?

First AP: a = 63, d = 2, aₙ = 63 + (n-1)2

Second AP: A = 3, D = 7, Aₙ = 3 + (n-1)7

Set them equal: 63 + 2(n-1) = 3 + 7(n-1)

60 + 2n - 2 = 7n - 7

58 + 2n = 7n - 7

65 = 5n ⇒ n = 13

The 13th terms of both APs are equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

a₃ = a + 2d = 16

a₇ - a₅ = (a + 6d) - (a + 4d) = 2d = 12 ⇒ d = 6

a = 16 - 2(6) = 4

The AP is: 4, 10, 16, 22, 28, ...

17. Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.

First, find the total number of terms:

a = 3, d = 5, aₙ = 253

253 = 3 + (n-1)5 ⇒ 250 = (n-1)5 ⇒ n-1 = 50 ⇒ n = 51

The 20th term from the last is the (51 - 20 + 1) = 32nd term from the beginning.

a₃₂ = 3 + 31(5) = 3 + 155 = 158

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

a₄ + a₈ = (a + 3d) + (a + 7d) = 2a + 10d = 24

a₆ + a₁₀ = (a + 5d) + (a + 9d) = 2a + 14d = 44

Subtracting: 4d = 20 ⇒ d = 5

2a + 10(5) = 24 ⇒ 2a + 50 = 24 ⇒ 2a = -26 ⇒ a = -13

The first three terms are: -13, -8, -3

19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

This forms an AP: 5000, 5200, 5400, ..., 7000

a = 5000, d = 200, aₙ = 7000

7000 = 5000 + (n-1)200 ⇒ 2000 = (n-1)200 ⇒ n-1 = 10 ⇒ n = 11

His income reached ₹7000 in the 11th year, which is 2005.

20. Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.

This forms an AP: 5, 6.75, 8.50, ..., 20.75

a = 5, d = 1.75, aₙ = 20.75

20.75 = 5 + (n-1)1.75 ⇒ 15.75 = (n-1)1.75 ⇒ n-1 = 9 ⇒ n = 10

Her savings became ₹20.75 in the 10th week.

Exercise 5.3 Solutions

1. Find the sum of the following APs:

(i) 2, 7, 12, ..., to 10 terms

a = 2, d = 5, n = 10

S₁₀ = 10/2 [2×2 + (10-1)5] = 5[4 + 45] = 5×49 = 245

(ii) -37, -33, -29, ..., to 12 terms

a = -37, d = 4, n = 12

S₁₂ = 12/2 [2×(-37) + (12-1)4] = 6[-74 + 44] = 6×(-30) = -180

(iii) 0.6, 1.7, 2.8, ..., to 100 terms

a = 0.6, d = 1.1, n = 100

S₁₀₀ = 100/2 [2×0.6 + (100-1)1.1] = 50[1.2 + 108.9] = 50×110.1 = 5505

(iv) 1/15, 1/12, 1/10, ..., to 11 terms

a = 1/15, d = 1/12 - 1/15 = (5-4)/60 = 1/60, n = 11

S₁₁ = 11/2 [2×(1/15) + (11-1)(1/60)] = 11/2 [2/15 + 10/60] = 11/2 [2/15 + 1/6]

= 11/2 [(8+5)/30] = 11/2 × 13/30 = 143/60

2. Find the sums given below:

(i) 7 + 10½ + 14 + ... + 84

a = 7, d = 3.5, aₙ = 84

84 = 7 + (n-1)3.5 ⇒ 77 = (n-1)3.5 ⇒ n-1 = 22 ⇒ n = 23

S₂₃ = 23/2 [7 + 84] = 23/2 × 91 = 2093/2 = 1046.5

(ii) 34 + 32 + 30 + ... + 10

a = 34, d = -2, aₙ = 10

10 = 34 + (n-1)(-2) ⇒ -24 = (n-1)(-2) ⇒ n-1 = 12 ⇒ n = 13

S₁₃ = 13/2 [34 + 10] = 13/2 × 44 = 13 × 22 = 286

(iii) -5 + (-8) + (-11) + ... + (-230)

a = -5, d = -3, aₙ = -230

-230 = -5 + (n-1)(-3) ⇒ -225 = (n-1)(-3) ⇒ n-1 = 75 ⇒ n = 76

S₇₆ = 76/2 [-5 + (-230)] = 38 × (-235) = -8930

3. In an AP:

(i) Given a = 5, d = 3, aₙ = 50, find n and Sₙ.

50 = 5 + (n-1)3 ⇒ 45 = (n-1)3 ⇒ n-1 = 15 ⇒ n = 16

S₁₆ = 16/2 [5 + 50] = 8 × 55 = 440

(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.

35 = 7 + 12d ⇒ 28 = 12d ⇒ d = 7/3

S₁₃ = 13/2 [7 + 35] = 13/2 × 42 = 13 × 21 = 273

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.

37 = a + 11×3 ⇒ 37 = a + 33 ⇒ a = 4

S₁₂ = 12/2 [4 + 37] = 6 × 41 = 246

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.

a₃ = a + 2d = 15

S₁₀ = 10/2 [2a + 9d] = 5[2a + 9d] = 125 ⇒ 2a + 9d = 25

From a + 2d = 15 ⇒ a = 15 - 2d

Substitute: 2(15 - 2d) + 9d = 25 ⇒ 30 - 4d + 9d = 25 ⇒ 30 + 5d = 25 ⇒ 5d = -5 ⇒ d = -1

a = 15 - 2(-1) = 17

a₁₀ = 17 + 9(-1) = 8

(v) Given d = 5, S₉ = 75, find a and a₉.

S₉ = 9/2 [2a + 8×5] = 9/2 [2a + 40] = 75

9/2 [2a + 40] = 75 ⇒ 9[2a + 40] = 150 ⇒ 18a + 360 = 150 ⇒ 18a = -210 ⇒ a = -35/3

a₉ = -35/3 + 8×5 = -35/3 + 40 = (-35 + 120)/3 = 85/3

(vi) Given a = 2, d = 8, Sₙ = 90, find n and aₙ.

Sₙ = n/2 [2×2 + (n-1)8] = n/2 [4 + 8n - 8] = n/2 [8n - 4] = n(4n - 2) = 90

4n² - 2n - 90 = 0 ⇒ 2n² - n - 45 = 0 ⇒ (2n + 9)(n - 5) = 0 ⇒ n = 5 (positive integer)

a₅ = 2 + 4×8 = 34

(vii) Given a = 8, aₙ = 62, Sₙ = 210, find n and d.

Sₙ = n/2 [8 + 62] = n/2 × 70 = 35n = 210 ⇒ n = 6

62 = 8 + 5d ⇒ 54 = 5d ⇒ d = 10.8

(viii) Given aₙ = 4, d = 2, Sₙ = -14, find n and a.

4 = a + (n-1)2 ⇒ a = 4 - 2(n-1) = 6 - 2n

Sₙ = n/2 [a + 4] = n/2 [6 - 2n + 4] = n/2 [10 - 2n] = n(5 - n) = -14

5n - n² = -14 ⇒ n² - 5n - 14 = 0 ⇒ (n - 7)(n + 2) = 0 ⇒ n = 7

a = 6 - 2×7 = -8

(ix) Given a = 3, n = 8, S = 192, find d.

192 = 8/2 [2×3 + 7d] = 4[6 + 7d] ⇒ 6 + 7d = 48 ⇒ 7d = 42 ⇒ d = 6

(x) Given l = 28, S = 144, and there are 9 terms, find a.

144 = 9/2 [a + 28] ⇒ 16 = (a + 28)/2 ⇒ a + 28 = 32 ⇒ a = 4

4. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?

a = 9, d = 8, Sₙ = 636

636 = n/2 [2×9 + (n-1)8] = n/2 [18 + 8n - 8] = n/2 [8n + 10] = n(4n + 5)

4n² + 5n - 636 = 0

Discriminant = 25 + 4×4×636 = 25 + 10176 = 10201

√10201 = 101

n = (-5 ± 101)/(2×4) = (-5 + 101)/8 = 96/8 = 12 or (-5 - 101)/8 = -106/8 (reject)

12 terms must be taken.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

a = 5, l = 45, Sₙ = 400

400 = n/2 [5 + 45] = n/2 × 50 = 25n ⇒ n = 16

45 = 5 + 15d ⇒ 40 = 15d ⇒ d = 8/3

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

a = 17, l = 350, d = 9

350 = 17 + (n-1)9 ⇒ 333 = (n-1)9 ⇒ n-1 = 37 ⇒ n = 38

S₃₈ = 38/2 [17 + 350] = 19 × 367 = 6973

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

a₂₂ = a + 21×7 = 149 ⇒ a + 147 = 149 ⇒ a = 2

S₂₂ = 22/2 [2 + 149] = 11 × 151 = 1661

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

a₂ = a + d = 14

a₃ = a + 2d = 18

Subtracting: d = 4

a = 14 - 4 = 10

S₅₁ = 51/2 [2×10 + 50×4] = 51/2 [20 + 200] = 51/2 × 220 = 51 × 110 = 5610

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

S₇ = 7/2 [2a + 6d] = 7(a + 3d) = 49 ⇒ a + 3d = 7

S₁₇ = 17/2 [2a + 16d] = 17(a + 8d) = 289 ⇒ a + 8d = 17

Subtracting: 5d = 10 ⇒ d = 2

a = 7 - 3×2 = 1

Sₙ = n/2 [2×1 + (n-1)2] = n/2 [2 + 2n - 2] = n/2 × 2n = n²

10. Show that a₁, a₂, ..., aₙ, ... form an AP where aₙ is defined as below:

(i) aₙ = 3 + 4n

aₙ = 4n + 3

aₙ₊₁ = 4(n+1) + 3 = 4n + 7

aₙ₊₁ - aₙ = (4n + 7) - (4n + 3) = 4 (constant)

So it forms an AP with d = 4.

(ii) aₙ = 9 - 5n

aₙ = -5n + 9

aₙ₊₁ = -5(n+1) + 9 = -5n + 4

aₙ₊₁ - aₙ = (-5n + 4) - (-5n + 9) = -5 (constant)

So it forms an AP with d = -5.

11. If the sum of the first n terms of an AP is 4n - n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Sₙ = 4n - n²

First term a₁ = S₁ = 4×1 - 1² = 3

S₂ = 4×2 - 2² = 8 - 4 = 4

Second term a₂ = S₂ - S₁ = 4 - 3 = 1

Third term a₃ = S₃ - S₂ = (12 - 9) - 4 = 3 - 4 = -1

We can find d = a₂ - a₁ = 1 - 3 = -2

10th term a₁₀ = 3 + 9×(-2) = 3 - 18 = -15

nth term aₙ = 3 + (n-1)(-2) = 3 - 2n + 2 = 5 - 2n

12. Find the sum of the first 40 positive integers divisible by 6.

The AP is: 6, 12, 18, ..., up to 40 terms

a = 6, d = 6, n = 40

S₄₀ = 40/2 [2×6 + 39×6] = 20[12 + 234] = 20×246 = 4920

13. Find the sum of the first 15 multiples of 8.

The AP is: 8, 16, 24, ..., up to 15 terms

a = 8, d = 8, n = 15

S₁₅ = 15/2 [2×8 + 14×8] = 15/2 [16 + 112] = 15/2 × 128 = 15 × 64 = 960

14. Find the sum of the odd numbers between 0 and 50.

The odd numbers between 0 and 50: 1, 3, 5, ..., 49

This is an AP with a = 1, l = 49, d = 2

49 = 1 + (n-1)2 ⇒ 48 = (n-1)2 ⇒ n-1 = 24 ⇒ n = 25

S₂₅ = 25/2 [1 + 49] = 25/2 × 50 = 25 × 25 = 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

This forms an AP: 200, 250, 300, ... with 30 terms

a = 200, d = 50, n = 30

S₃₀ = 30/2 [2×200 + 29×50] = 15[400 + 1450] = 15×1850 = 27750

The contractor has to pay ₹27,750 as penalty.

16. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Let the first prize be ₹a

Then the prizes form an AP: a, a-20, a-40, ..., with 7 terms

S₇ = 7/2 [2a + 6×(-20)] = 7/2 [2a - 120] = 7(a - 60) = 700

a - 60 = 100 ⇒ a = 160

The prizes are: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Each class from I to XII has 3 sections, and each section plants trees equal to their class number.

So trees planted by: Class I: 3×1 = 3, Class II: 3×2 = 6, ..., Class XII: 3×12 = 36

This forms an AP: 3, 6, 9, ..., 36

a = 3, d = 3, l = 36

36 = 3 + (n-1)3 ⇒ 33 = (n-1)3 ⇒ n-1 = 11 ⇒ n = 12

S₁₂ = 12/2 [3 + 36] = 6 × 39 = 234

Total trees planted = 234

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Length of a semicircle = πr

So the lengths form an AP: π×0.5, π×1.0, π×1.5, ..., π×6.5

This is: 0.5π, π, 1.5π, ..., 6.5π

a = 0.5π, d = 0.5π, n = 13

S₁₃ = 13/2 [2×0.5π + 12×0.5π] = 13/2 [π + 6π] = 13/2 × 7π = 13/2 × 7 × 22/7 = 13 × 11 = 143 cm

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?

This forms an AP: 20, 19, 18, ...

a = 20, d = -1, Sₙ = 200

200 = n/2 [2×20 + (n-1)(-1)] = n/2 [40 - n + 1] = n/2 [41 - n]

400 = n(41 - n) ⇒ n² - 41n + 400 = 0

Discriminant = 1681 - 1600 = 81

n = (41 ± 9)/2 = 25 or 16

Both are valid, but typically we'd use n = 16 (since with n = 25, we'd have negative terms)

With n = 16: a₁₆ = 20 + 15×(-1) = 5

There are 16 rows with 5 logs in the top row.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

For the first potato: distance = 5 + 5 = 10 m

For the second potato: distance = (5+3) + (5+3) = 16 m

For the third potato: distance = (5+6) + (5+6) = 22 m

This forms an AP: 10, 16, 22, ... with 10 terms

a = 10, d = 6, n = 10

S₁₀ = 10/2 [2×10 + 9×6] = 5[20 + 54] = 5×74 = 370 m

The competitor has to run 370 m.

Exercise 5.4 Solutions (Optional)

1. Which term of the AP: 121, 117, 113, ..., is its first negative term?

a = 121, d = -4

We need aₙ < 0

121 + (n-1)(-4) < 0 ⇒ 121 - 4n + 4 < 0 ⇒ 125 < 4n ⇒ n > 31.25

The smallest integer greater than 31.25 is 32.

The 32nd term is the first negative term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

a₃ + a₇ = (a + 2d) + (a + 6d) = 2a + 8d = 6 ⇒ a + 4d = 3

a₃ × a₇ = (a + 2d)(a + 6d) = 8

From a + 4d = 3 ⇒ a = 3 - 4d

Substitute: (3 - 4d + 2d)(3 - 4d + 6d) = (3 - 2d)(3 + 2d) = 9 - 4d² = 8

4d² = 1 ⇒ d² = 1/4 ⇒ d = ±1/2

Case 1: d = 1/2, a = 3 - 4(1/2) = 1

S₁₆ = 16/2 [2×1 + 15×1/2] = 8[2 + 7.5] = 8×9.5 = 76

Case 2: d = -1/2, a = 3 - 4(-1/2) = 5

S₁₆ = 16/2 [2×5 + 15×(-1/2)] = 8[10 - 7.5] = 8×2.5 = 20

3. A ladder has rungs 25 cm apart (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs?

Distance between top and bottom rungs = 2.5 m = 250 cm

Rungs are 25 cm apart, so number of rungs = 250/25 + 1 = 10 + 1 = 11

Lengths of rungs form an AP: 45, ?, ..., 25

a = 45, l = 25, n = 11

25 = 45 + 10d ⇒ -20 = 10d ⇒ d = -2

S₁₁ = 11/2 [45 + 25] = 11/2 × 70 = 11 × 35 = 385 cm

Length of wood required = 385 cm = 3.85 m

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

House numbers: 1, 2, 3, ..., 49

Sum of numbers before x = Sum of numbers after x

Sum of first (x-1) terms = Sum from (x+1) to 49

(x-1)/2 [1 + (x-1)] = (49-x)/2 [(x+1) + 49]

(x-1)/2 × x = (49-x)/2 × (x + 50)

x(x-1) = (49-x)(x+50)

x² - x = 49x + 2450 - x² - 50x

x² - x = -x² - x + 2450

2x² = 2450

x² = 1225

x = 35 (since x must be between 1 and 49)

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of ¼ m and a tread of ½ m (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.

Volume of step 1 = length × width × height = 50 × 1/2 × 1/4 = 50/8 = 6.25 m³

Volume of step 2 = 50 × 1/2 × (1/4 + 1/4) = 50 × 1/2 × 1/2 = 12.5 m³

Volume of step 3 = 50 × 1/2 × (1/4 + 1/4 + 1/4) = 50 × 1/2 × 3/4 = 18.75 m³

This forms an AP: 6.25, 12.5, 18.75, ... with 15 terms

a = 6.25, d = 6.25, n = 15

S₁₅ = 15/2 [2×6.25 + 14×6.25] = 15/2 [12.5 + 87.5] = 15/2 × 100 = 15 × 50 = 750 m³

Total volume of concrete required = 750 m³

Key Formulas for Arithmetic Progressions

General Form of an AP

a, a+d, a+2d, a+3d, ..., a+(n-1)d

nth Term of an AP

aₙ = a + (n-1)d

Sum of First n Terms of an AP

Sₙ = n/2 [2a + (n-1)d]

Sₙ = n/2 [a + l] where l is the last term

Common Difference

d = a₂ - a₁ = a₃ - a₂ = ... = aₙ - aₙ₋₁

Relation Between Three Consecutive Terms

If a, b, c are in AP, then 2b = a + c

Important Terms

Arithmetic Progression (AP): A sequence of numbers in which the difference between consecutive terms is constant.
First Term (a): The initial term of the AP.
Common Difference (d): The constant difference between consecutive terms.
nth Term (aₙ): The term at position n in the AP.
Sum of First n Terms (Sₙ): The sum of the first n terms of the AP.