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NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

Arithmetic Progressions explains number patterns where every next term is formed by adding a fixed difference.
These NCERT Solutions help students solve Chapter 5 questions on AP terms, common difference, nth term and sums.

Chapter 5 Arithmetic Progressions begins with patterns from salary increments, ladder rungs, savings and rows of plants. It then teaches students to check whether a list is an AP, find the nth term, and calculate the sum of terms. NCERT Solutions Class 10 Maths Chapter 5 cover Exercise 5.1, Exercise 5.2, Exercise 5.3 and optional Exercise 5.4. Students practise common difference, missing terms, word problems, salary growth, penalties, tree planting and stacked logs for 2026-27 CBSE exams.

Key Takeaways

Arithmetic progression: Each term is formed by adding a fixed common difference.
nth term: The formula aₙ = a + (n - 1)d helps find any term.
Sum of n terms: The formula Sₙ = n / 2 [2a + (n - 1)d] gives total sum.
Word problems: Salary, savings, penalties, trees and logs use AP patterns.

NCERT Solutions Class 10 Maths Chapter 5 Structure 2026-27

Exercise	Topic	Question Count
Exercise 5.1	Identifying APs and common difference	4
Exercise 5.2	nth term of an AP	20
Exercise 5.3	Sum of first n terms	20
Exercise 5.4	Optional AP applications	5

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise

The NCERT chapter has four exercises on AP patterns, nth term and sums. These Arithmetic Progressions Class 10 questions and answers follow the textbook order.

Class 10 Maths Chapter 5 Exercise 5.1 Solutions

Exercise 5.1 checks whether a sequence forms an AP. Students mainly use the common difference between consecutive terms.

Q1. In which situations does the list of numbers make an arithmetic progression?

(i) Taxi fare after each km

Answer: Yes, it forms an AP.

The fares are:

15, 23, 31, 39, ...

Each term increases by 8.

Common difference = 8

(ii) Air remaining in a cylinder

Answer: No, it does not form an AP.

The pump removes 1 / 4 of the remaining air each time.

The remaining air is multiplied by a fixed fraction.

So, the difference is not constant.

(iii) Cost of digging a well after every metre

Answer: Yes, it forms an AP.

The costs are:

150, 200, 250, 300, ...

Each metre costs Rs 50 more than the previous metre.

Common difference = 50

(iv) Amount at compound interest

Answer: No, it does not form an AP.

The amount increases by a percentage each year.

The difference between terms is not constant.

Q2. Write the first four terms of the AP.

(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

Q3. For the following APs, write the first term and common difference.

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

First term = 3

Common difference = 1 - 3 = -2

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

First term = -5

Common difference = -1 - (-5) = 4

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

First term = 1 / 3

Common difference = 5 / 3 - 1 / 3 = 4 / 3

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

First term = 0.6

Common difference = 1.7 - 0.6 = 1.1

Q4. Which of the following are APs?

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

Answer: Not an AP.

The differences are 2, 4 and 8.

They are not equal.

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

Answer: It is an AP.

Common difference = 1 / 2

Next three terms:

4, 9 / 2, 5

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

Answer: It is an AP.

Common difference = -2

Next three terms:

-9.2, -11.2, -13.2

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

Answer: It is an AP.

Common difference = 4

Next three terms:

6, 10, 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ...

Answer: It is an AP.

Common difference = √2

Next three terms:

3 + 4√2, 3 + 5√2, 3 + 6√2

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

Answer: Not an AP.

The differences are not equal.

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

Answer: It is an AP.

Common difference = -4

Next three terms:

-16, -20, -24

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

Answer: It is an AP.

Common difference = 0

Next three terms:

-1 / 2, -1 / 2, -1 / 2

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

Answer: Not an AP.

The differences are 2, 6 and 18.

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

Answer: It is an AP.

Common difference = a

Next three terms:

5a, 6a, 7a

(xi) a, a², a³, a⁴, ...

Answer: It is not always an AP.

The differences depend on the value of a.

(xii) 2, 8, 18, 32, ...

Answer: Not an AP.

The differences are 6, 10 and 14.

(xiii) √3, √6, √9, √12, ...

Answer: Not an AP.

The differences are not equal.

(xiv) 1², 3², 5², 7², ...

Answer: Not an AP.

The terms are 1, 9, 25 and 49.

The differences are not equal.

(xv) 1², 5², 7², 7³, ...

Answer: Not an AP.

The terms do not have a constant difference.

Class 10 Maths Chapter 5 Exercise 5.2 Solutions

Exercise 5.2 uses the nth term formula. This part of Class 10 Maths Chapter 5 Arithmetic Progressions helps students find missing terms and positions.

Q1. Fill in the blanks in the table.

Formula:

aₙ = a + (n - 1)d

Part	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

Q2. Choose the correct answer.

(i) 30th term of 10, 7, 4, ...

Given:
a = 10
d = -3
n = 30

Formula:

aₙ = a + (n - 1)d

a₃₀ = 10 + 29(-3)

a₃₀ = 10 - 87

a₃₀ = -77

Answer: Option (C) -77

(ii) 11th term of -3, -1 / 2, 2, ...

Given:
a = -3
d = 5 / 2
n = 11

a₁₁ = -3 + 10(5 / 2)

a₁₁ = -3 + 25

a₁₁ = 22

Answer: Option (B) 22

Q3. Find the missing terms.

(i) 2, __, 26

There are three terms.

Middle term = (2 + 26) / 2

Middle term = 14

Answer: 2, 14, 26

(ii) , 13, , 3

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

Common difference = -5

First term = 18

Third term = 8

Answer: 18, 13, 8, 3

(iii) 5, , , 9½

Common difference = (9.5 - 5) / 3

Common difference = 1.5

Missing terms are:

6.5 and 8

Answer: 5, 6.5, 8, 9.5

(iv) -4, , , , , 6

Common difference = (6 - (-4)) / 5

Common difference = 2

Answer: -4, -2, 0, 2, 4, 6

(v) , 38, , , , -22

Common difference = -15

First term = 53

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

Q4. Which term of 3, 8, 13, 18, ... is 78?

Given:
a = 3
d = 5
aₙ = 78

Formula:

aₙ = a + (n - 1)d

78 = 3 + (n - 1)5

75 = 5(n - 1)

n - 1 = 15

n = 16

Final Answer:
78 is the 16th term.

Q5. Find the number of terms.

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

Given:
a = 7
d = 6
aₙ = 205

205 = 7 + (n - 1)6

198 = 6(n - 1)

n - 1 = 33

n = 34

Final Answer:
There are 34 terms.

(ii) 18, 15½, 13, ..., -47

Given:
a = 18
d = -2.5
aₙ = -47

-47 = 18 + (n - 1)(-2.5)

-65 = -2.5(n - 1)

n - 1 = 26

n = 27

Final Answer:
There are 27 terms.

Q6. Check whether -150 is a term of 11, 8, 5, 2, ...

Given:
a = 11
d = -3
aₙ = -150

-150 = 11 + (n - 1)(-3)

-161 = -3(n - 1)

n - 1 = 161 / 3

n is not a positive integer.

Final Answer:
-150 is not a term of the AP.

Q7. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Given:
a₁₁ = 38
a₁₆ = 73

a₁₆ - a₁₁ = 5d

73 - 38 = 5d

35 = 5d

d = 7

Now:

a₁₁ = a + 10d

38 = a + 70

a = -32

a₃₁ = a + 30d

a₃₁ = -32 + 210

a₃₁ = 178

Final Answer:
The 31st term is 178.

Q8. An AP has 50 terms. Its 3rd term is 12 and last term is 106. Find the 29th term.

Given:
a₃ = 12
a₅₀ = 106

a + 2d = 12

a + 49d = 106

Subtract:

47d = 94

d = 2

a + 4 = 12

a = 8

a₂₉ = a + 28d

a₂₉ = 8 + 56

a₂₉ = 64

Final Answer:
The 29th term is 64.

Q9. If the 3rd and 9th terms are 4 and -8, which term is zero?

Given:
a₃ = 4
a₉ = -8

a + 2d = 4

a + 8d = -8

Subtract:

6d = -12

d = -2

a + 2(-2) = 4

a = 8

For zero term:

0 = 8 + (n - 1)(-2)

2(n - 1) = 8

n - 1 = 4

n = 5

Final Answer:
The 5th term is zero.

Q10. The 17th term exceeds the 10th term by 7. Find d.

a₁₇ - a₁₀ = 7

[a + 16d] - [a + 9d] = 7

7d = 7

d = 1

Final Answer:
The common difference is 1.

Q11. Which term of 3, 15, 27, 39, ... is 132 more than its 54th term?

Given:
d = 12

Let the required term be nth term.

aₙ - a₅₄ = 132

(n - 54)d = 132

(n - 54)12 = 132

n - 54 = 11

n = 65

Final Answer:
The 65th term is 132 more than the 54th term.

Q12. Two APs have the same common difference. Their 100th terms differ by 100. Find the difference between their 1000th terms.

If two APs have the same common difference, the difference between their nth terms remains constant.

Difference between 100th terms = 100

So, difference between 1000th terms = 100

Final Answer:
The difference is 100.

Q13. How many three-digit numbers are divisible by 7?

Smallest three-digit multiple of 7 = 105

Largest three-digit multiple of 7 = 994

AP:

105, 112, 119, ..., 994

Given:
a = 105
d = 7
aₙ = 994

994 = 105 + (n - 1)7

889 = 7(n - 1)

n - 1 = 127

n = 128

Final Answer:
There are 128 three-digit numbers divisible by 7.

Q14. How many multiples of 4 lie between 10 and 250?

Smallest multiple = 12

Largest multiple = 248

AP:

12, 16, 20, ..., 248

Given:
a = 12
d = 4
aₙ = 248

248 = 12 + (n - 1)4

236 = 4(n - 1)

n - 1 = 59

n = 60

Final Answer:
There are 60 multiples of 4.

Q15. For what value of n are nth terms of two APs equal?

AP 1:

63, 65, 67, ...

a = 63, d = 2

nth term = 63 + (n - 1)2

nth term = 61 + 2n

AP 2:

3, 10, 17, ...

a = 3, d = 7

nth term = 3 + (n - 1)7

nth term = 7n - 4

Set them equal:

61 + 2n = 7n - 4

65 = 5n

n = 13

Final Answer:
The nth terms are equal when n = 13.

Q16. Determine the AP whose third term is 16 and 7th term exceeds 5th term by 12.

Given:
a₃ = 16

a + 2d = 16

Also:

a₇ - a₅ = 12

(a + 6d) - (a + 4d) = 12

2d = 12

d = 6

a + 2(6) = 16

a = 4

Final Answer:
The AP is 4, 10, 16, 22, ...

Q17. Find the 20th term from the last term of 3, 8, 13, ..., 253.

Given:
a = 3
d = 5
last term = 253

First, find total terms.

253 = 3 + (n - 1)5

250 = 5(n - 1)

n - 1 = 50

n = 51

20th term from the last = 32nd term from the start

a₃₂ = 3 + 31(5)

a₃₂ = 158

Final Answer:
The 20th term from the last is 158.

Q18. Sum of 4th and 8th terms is 24. Sum of 6th and 10th terms is 44. Find first three terms.

a₄ + a₈ = 24

(a + 3d) + (a + 7d) = 24

2a + 10d = 24

a + 5d = 12

Now:

a₆ + a₁₀ = 44

(a + 5d) + (a + 9d) = 44

2a + 14d = 44

a + 7d = 22

Subtract:

2d = 10

d = 5

a + 25 = 12

a = -13

First three terms:

-13, -8, -3

Final Answer:
The first three terms are -13, -8, -3.

Q19. Subba Rao started work in 1995 at Rs 5000 salary. Increment is Rs 200 each year. In which year did income reach Rs 7000?

Given:
a = 5000
d = 200
aₙ = 7000

7000 = 5000 + (n - 1)200

2000 = 200(n - 1)

n - 1 = 10

n = 11

The first year is 1995.

11th year = 2005

Final Answer:
His income reached Rs 7000 in 2005.

Q20. Ramkali saved Rs 5 in first week and increased weekly savings by Rs 1.75. In nth week, savings became Rs 20.75. Find n.

Given:
a = 5
d = 1.75
aₙ = 20.75

20.75 = 5 + (n - 1)1.75

15.75 = 1.75(n - 1)

n - 1 = 9

n = 10

Final Answer:
n = 10

Class 10 Maths Chapter 5 Exercise 5.3 Solutions

Exercise 5.3 uses AP sum formulas. Class 10 Maths Chapter 5 Exercise 5.3 is important for word problems and long calculations.

Q1. Find the sum of the following APs.

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

Given:
a = 2
d = 5
n = 10

Formula:

Sₙ = n / 2 [2a + (n - 1)d]

S₁₀ = 10 / 2 [4 + 9(5)]

S₁₀ = 5 [4 + 45]

S₁₀ = 245

Final Answer:
Sum = 245

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

Given:
a = -37
d = 4
n = 12

S₁₂ = 12 / 2 [2(-37) + 11(4)]

S₁₂ = 6 [-74 + 44]

S₁₂ = -180

Final Answer:
Sum = -180

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

Given:
a = 0.6
d = 1.1
n = 100

S₁₀₀ = 100 / 2 [2(0.6) + 99(1.1)]

S₁₀₀ = 50 [1.2 + 108.9]

S₁₀₀ = 5505

Final Answer:
Sum = 5505

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

Given:
a = 1 / 15
d = 1 / 60
n = 11

S₁₁ = 11 / 2 [2 / 15 + 10 / 60]

S₁₁ = 11 / 2 [8 / 60 + 10 / 60]

S₁₁ = 11 / 2 × 18 / 60

S₁₁ = 33 / 20

Final Answer:
Sum = 33 / 20

Q2. Find the sums.

(i) 7 + 10½ + 14 + ... + 84

Given:
a = 7
d = 3.5
last term = 84

Find n:

84 = 7 + (n - 1)3.5

77 = 3.5(n - 1)

n - 1 = 22

n = 23

Now:

S = n / 2 (a + l)

S = 23 / 2 (7 + 84)

S = 23 / 2 × 91

S = 2093 / 2

Final Answer:
Sum = 2093 / 2

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

Given:
a = 34
d = -2
l = 10

10 = 34 + (n - 1)(-2)

-24 = -2(n - 1)

n = 13

S = 13 / 2 (34 + 10)

S = 286

Final Answer:
Sum = 286

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

Given:
a = -5
d = -3
l = -230

-230 = -5 + (n - 1)(-3)

-225 = -3(n - 1)

n = 76

S = 76 / 2 (-5 - 230)

S = 38(-235)

S = -8930

Final Answer:
Sum = -8930

Q3. Find the missing values.

(i) a = 5, d = 3, aₙ = 50

50 = 5 + (n - 1)3

45 = 3(n - 1)

n = 16

Sₙ = 16 / 2 (5 + 50)

Sₙ = 440

Final Answer:
n = 16, Sₙ = 440

(ii) a = 7, a₁₃ = 35

35 = 7 + 12d

28 = 12d

d = 7 / 3

S₁₃ = 13 / 2 (7 + 35)

S₁₃ = 273

Final Answer:
d = 7 / 3, S₁₃ = 273

(iii) a₁₂ = 37, d = 3

37 = a + 11(3)

37 = a + 33

a = 4

S₁₂ = 12 / 2 (4 + 37)

S₁₂ = 246

Final Answer:
a = 4, S₁₂ = 246

(iv) a₃ = 15, S₁₀ = 125

a + 2d = 15

S₁₀ = 10 / 2 [2a + 9d]

125 = 5[2a + 9d]

2a + 9d = 25

Using a = 15 - 2d:

2(15 - 2d) + 9d = 25

30 - 4d + 9d = 25

5d = -5

d = -1

a = 17

a₁₀ = 17 + 9(-1)

a₁₀ = 8

Final Answer:
d = -1, a₁₀ = 8

(v) d = 5, S₉ = 75

S₉ = 9 / 2 [2a + 8d]

75 = 9 / 2 [2a + 40]

150 = 9[2a + 40]

2a + 40 = 50 / 3

2a = -70 / 3

a = -35 / 3

a₉ = a + 8d

a₉ = -35 / 3 + 40

a₉ = 85 / 3

Final Answer:
a = -35 / 3, a₉ = 85 / 3

(vi) a = 2, d = 8, Sₙ = 90

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

90 = n / 2 [4 + 8n - 8]

90 = n / 2 [8n - 4]

90 = n(4n - 2)

4n² - 2n - 90 = 0

2n² - n - 45 = 0

n = 5

aₙ = 2 + 4(8)

aₙ = 34

Final Answer:
n = 5, aₙ = 34

(vii) a = 8, aₙ = 62, Sₙ = 210

Sₙ = n / 2 (a + aₙ)

210 = n / 2 (8 + 62)

210 = 35n

n = 6

62 = 8 + 5d

54 = 5d

d = 54 / 5

Final Answer:
n = 6, d = 54 / 5

(viii) aₙ = 4, d = 2, Sₙ = -14

a = aₙ - (n - 1)d

a = 4 - 2(n - 1)

a = 6 - 2n

Sₙ = n / 2 (a + aₙ)

-14 = n / 2 (6 - 2n + 4)

-14 = n / 2 (10 - 2n)

-14 = n(5 - n)

n² - 5n - 14 = 0

n = 7

a = 6 - 14

a = -8

Final Answer:
n = 7, a = -8

(ix) a = 3, n = 8, S = 192

192 = 8 / 2 [2(3) + 7d]

192 = 4[6 + 7d]

48 = 6 + 7d

7d = 42

d = 6

Final Answer:
d = 6

(x) l = 28, S = 144, n = 9

S = n / 2 (a + l)

144 = 9 / 2 (a + 28)

288 = 9(a + 28)

32 = a + 28

a = 4

Final Answer:
a = 4

Q4. How many terms of 9, 17, 25, ... give sum 636?

Given:
a = 9
d = 8
Sₙ = 636

636 = n / 2 [18 + (n - 1)8]

636 = n / 2 [8n + 10]

1272 = n(8n + 10)

4n² + 5n - 636 = 0

n = 12

Final Answer:
12 terms must be taken.

Q5. First term is 5, last term is 45 and sum is 400. Find n and d.

Given:
a = 5
l = 45
S = 400

S = n / 2 (a + l)

400 = n / 2 (50)

400 = 25n

n = 16

Now:

l = a + (n - 1)d

45 = 5 + 15d

40 = 15d

d = 8 / 3

Final Answer:
n = 16, d = 8 / 3

Q6. First and last terms are 17 and 350. Common difference is 9.

Given:
a = 17
l = 350
d = 9

Find n:

350 = 17 + (n - 1)9

333 = 9(n - 1)

n - 1 = 37

n = 38

Find sum:

S = n / 2 (a + l)

S = 38 / 2 (17 + 350)

S = 19 × 367

S = 6973

Final Answer:
There are 38 terms and their sum is 6973.

Q7. Find sum of first 22 terms when d = 7 and 22nd term is 149.

Given:
d = 7
a₂₂ = 149

Find a:

149 = a + 21(7)

149 = a + 147

a = 2

Now:

S₂₂ = 22 / 2 (2 + 149)

S₂₂ = 11 × 151

S₂₂ = 1661

Final Answer:
Sum = 1661

Q8. Find sum of first 51 terms whose second and third terms are 14 and 18.

Given:
a₂ = 14
a₃ = 18

d = 18 - 14

d = 4

a + d = 14

a = 10

S₅₁ = 51 / 2 [2(10) + 50(4)]

S₅₁ = 51 / 2 [220]

S₅₁ = 5610

Final Answer:
Sum = 5610

Q9. If sum of first 7 terms is 49 and first 17 terms is 289, find Sₙ.

S₇ = 49

7 / 2 [2a + 6d] = 49

2a + 6d = 14

a + 3d = 7

S₁₇ = 289

17 / 2 [2a + 16d] = 289

2a + 16d = 34

a + 8d = 17

Subtract:

5d = 10

d = 2

a + 6 = 7

a = 1

Now:

Sₙ = n / 2 [2a + (n - 1)d]

Sₙ = n / 2 [2 + 2(n - 1)]

Sₙ = n / 2 [2n]

Sₙ = n²

Final Answer:
Sₙ = n²

Q10. Show that the sequence is an AP and find sum of first 15 terms.

(i) aₙ = 3 + 4n

a₁ = 7

a₂ = 11

a₃ = 15

The common difference is 4.

So, it is an AP.

S₁₅ = 15 / 2 (a₁ + a₁₅)

a₁₅ = 3 + 4(15)

a₁₅ = 63

S₁₅ = 15 / 2 (7 + 63)

S₁₅ = 525

Final Answer:
Sum = 525

(ii) aₙ = 9 - 5n

a₁ = 4

a₂ = -1

a₃ = -6

The common difference is -5.

So, it is an AP.

a₁₅ = 9 - 5(15)

a₁₅ = -66

S₁₅ = 15 / 2 (4 - 66)

S₁₅ = -465

Final Answer:
Sum = -465

Q11. If Sₙ = 4n - n², find terms.

S₁ = 4(1) - 1²

S₁ = 3

First term = 3

S₂ = 4(2) - 2²

S₂ = 4

Second term = S₂ - S₁

Second term = 4 - 3

Second term = 1

S₃ = 4(3) - 3²

S₃ = 3

Third term = S₃ - S₂

Third term = 3 - 4

Third term = -1

10th term:

a₁₀ = S₁₀ - S₉

S₁₀ = 40 - 100 = -60

S₉ = 36 - 81 = -45

a₁₀ = -60 - (-45)

a₁₀ = -15

nth term:

aₙ = Sₙ - Sₙ₋₁

aₙ = 5 - 2n

Final Answer:
First term = 3
Second term = 1
Third term = -1
10th term = -15
nth term = 5 - 2n

Q12. Find the sum of first 40 positive integers divisible by 6.

The numbers are:

6, 12, 18, ..., 240

Given:
a = 6
d = 6
n = 40

S₄₀ = 40 / 2 (6 + 240)

S₄₀ = 20 × 246

S₄₀ = 4920

Final Answer:
Sum = 4920

Q13. Find the sum of first 15 multiples of 8.

The numbers are:

8, 16, 24, ..., 120

S₁₅ = 15 / 2 (8 + 120)

S₁₅ = 15 / 2 × 128

S₁₅ = 960

Final Answer:
Sum = 960

Q14. Find the sum of odd numbers between 0 and 50.

Odd numbers are:

1, 3, 5, ..., 49

Given:
a = 1
l = 49
n = 25

S = 25 / 2 (1 + 49)

S = 25 / 2 × 50

S = 625

Final Answer:
Sum = 625

Q15. Penalty starts at Rs 200 and increases by Rs 50 each day. Find penalty for 30 days.

Given:
a = 200
d = 50
n = 30

S₃₀ = 30 / 2 [2(200) + 29(50)]

S₃₀ = 15 [400 + 1450]

S₃₀ = 15 × 1850

S₃₀ = 27750

Final Answer:
The contractor must pay Rs 27,750.

Q16. Rs 700 is divided into 7 prizes. Each prize is Rs 20 less than the previous prize.

Given:
n = 7
d = -20
S = 700

S₇ = 7 / 2 [2a + 6(-20)]

700 = 7 / 2 [2a - 120]

200 = 2a - 120

2a = 320

a = 160

Prizes:

160, 140, 120, 100, 80, 60, 40

Final Answer:
The prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60 and Rs 40.

Q17. Students plant trees from Class I to XII. There are three sections in each class.

Each section of Class I plants 1 tree.

Each section of Class II plants 2 trees.

This continues till Class XII.

Total trees per one section:

1 + 2 + 3 + ... + 12

S₁₂ = 12 / 2 (1 + 12)

S₁₂ = 78

There are three sections.

Total trees = 3 × 78

Total trees = 234

Final Answer:
Students will plant 234 trees.

Q18. Find the total length of a spiral with 13 semicircles.

Radii are:

0.5 cm, 1.0 cm, 1.5 cm, ..., 6.5 cm

Length of a semicircle = πr

Total length = π × sum of radii

Sum of radii:

S₁₃ = 13 / 2 (0.5 + 6.5)

S₁₃ = 13 / 2 × 7

S₁₃ = 45.5

Total length = 45.5π

Using π = 22 / 7:

Total length = 45.5 × 22 / 7

Total length = 143 cm

Final Answer:
Total length = 143 cm

Q19. 200 logs are stacked with 20 logs in bottom row, 19 in next row, and so on.

The rows form an AP:

20, 19, 18, ...

Given:
a = 20
d = -1
Sₙ = 200

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

200 = n / 2 [40 - n + 1]

400 = n(41 - n)

n² - 41n + 400 = 0

(n - 16)(n - 25) = 0

n = 16 or 25

n = 25 is not possible because top row would be negative.

So, n = 16.

Top row:

a₁₆ = 20 + 15(-1)

a₁₆ = 5

Final Answer:
There are 16 rows, and the top row has 5 logs.

Q20. Find total distance in the potato race.

Distances of potatoes from the bucket are:

5, 8, 11, ..., 32

There are 10 potatoes.

Each potato needs a trip to and from the bucket.

Total distance = 2 × sum of distances

Find sum:

S₁₀ = 10 / 2 [2(5) + 9(3)]

S₁₀ = 5 [10 + 27]

S₁₀ = 185

Total distance = 2 × 185

Total distance = 370 m

Final Answer:
The competitor runs 370 m.

Class 10 Maths Chapter 5 Exercise 5.4 Solutions

Exercise 5.4 is optional in the NCERT book. These answers are useful for extra AP practice.

Q1. Which term of 121, 117, 113, ... is the first negative term?

Given:
a = 121
d = -4

For first negative term:

aₙ < 0

121 + (n - 1)(-4) < 0

121 - 4n + 4 < 0

125 - 4n < 0

125 < 4n

n > 31.25

So, the first integer value is 32.

Final Answer:
The 32nd term is the first negative term.

Q2. Sum of 3rd and 7th terms is 6. Product is 8. Find sum of first 16 terms.

a₃ + a₇ = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

Also:

a₃ × a₇ = 8

(a + 2d)(a + 6d) = 8

Let a + 4d = 3.

Then:

(3 - 2d)(3 + 2d) = 8

9 - 4d² = 8

4d² = 1

d = 1 / 2 or -1 / 2

If d = 1 / 2:

a = 1

S₁₆ = 16 / 2 [2(1) + 15(1 / 2)]

S₁₆ = 76

If d = -1 / 2:

a = 5

S₁₆ = 16 / 2 [2(5) + 15(-1 / 2)]

S₁₆ = 20

Final Answer:
The sum is 76 or 20.

Q3. Ladder rungs are 25 cm apart. Length decreases from 45 cm to 25 cm.

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

Number of spaces = 250 / 25 = 10

Number of rungs = 10 + 1 = 11

Lengths form an AP.

Given:
a = 45
l = 25
n = 11

S = n / 2 (a + l)

S = 11 / 2 (45 + 25)

S = 11 / 2 × 70

S = 385 cm

Final Answer:
Length of wood required = 385 cm, or 3.85 m

Q4. Houses are numbered 1 to 49. Find x.

Sum before house x = Sum after house x

Total sum from 1 to 49:

S₄₉ = 49 / 2 (1 + 49)

S₄₉ = 1225

Sum before x:

1 + 2 + ... + (x - 1) = x(x - 1) / 2

Sum after x:

1225 - x(x + 1) / 2

Now:

x(x - 1) / 2 = 1225 - x(x + 1) / 2

x(x - 1) = 2450 - x(x + 1)

2x² = 2450

x² = 1225

x = 35

Final Answer:
The value of x is 35.

Q5. Find concrete volume for 15 steps.

Each step is 50 m long.

Rise = 1 / 4 m

Tread = 1 / 2 m

Volume of first step:

50 × 1 / 2 × 1 / 4 = 25 / 4 cubic metres

The volumes form an AP:

25 / 4, 2(25 / 4), 3(25 / 4), ..., 15(25 / 4)

Total volume:

25 / 4 × (1 + 2 + 3 + ... + 15)

Sum from 1 to 15:

15 / 2 (1 + 15) = 120

Total volume = 25 / 4 × 120

Total volume = 750 cubic metres

Final Answer:
Concrete required = 750 m³

Arithmetic Progressions Class 10 NCERT Solutions: Key Concepts

Arithmetic Progressions Class 10 NCERT Solutions depend on three values: first term, common difference and term number. Students should identify these before using any formula.

Arithmetic Progression

An AP is a list where each term is formed by adding a fixed number.

That fixed number is called the common difference.

Common Difference

Common difference = Second term - First term

If the same difference continues, the sequence is an AP.

nth Term

Formula:

aₙ = a + (n - 1)d

Use this when the question asks for a term or term position.

Sum of First n Terms

Formula:

Sₙ = n / 2 [2a + (n - 1)d]

Use this when first term, common difference and number of terms are known.

Sum Using Last Term

Formula:

Sₙ = n / 2 (a + l)

Use this when first term, last term and number of terms are known.

NCERT Solutions Class 10 Maths Chapter 5: Formula Use

NCERT Solutions Class 10 Maths Chapter 5 questions become easier when students select the right formula first. This table shows the best formula for each question type.

Question Type	Formula	Example Use
Check AP	d = next term - previous term	Exercise 5.1
Find nth term	aₙ = a + (n - 1)d	Exercise 5.2
Find number of terms	aₙ = a + (n - 1)d	Term position questions
Find sum	Sₙ = n / 2 [2a + (n - 1)d]	Exercise 5.3
Sum with last term	Sₙ = n / 2 (a + l)	Logs, prizes and finite APs

Salary and Increment Questions

Salary questions usually form an AP.

The starting salary is the first term, and annual increment is the common difference.

Rows and Arrangement Questions

Rows of plants, logs and ladder rungs often form finite APs.

The first row and last row help find n or total sum.

Savings and Penalty Questions

Savings and penalty questions usually ask for total amount.

Students should use the sum formula, not only the nth term formula.

Missing Term Questions

Missing terms need equal spacing between first and last terms.

Find the common difference, then fill terms one by one.

Optional AP Problems

Optional questions combine AP with geometry or conditions.

They should be solved slowly by forming the AP first.

Useful Links for Class 10 Maths NCERT Solutions

Section	Useful Links
Class 10 Maths NCERT Solutions	NCERT Solutions for Class 10 Maths
Chapter 1	NCERT Solutions for Class 10 Maths Chapter 1
Chapter 2	NCERT Solutions for Class 10 Maths Chapter 2
Chapter 3	NCERT Solutions for Class 10 Maths Chapter 3
Chapter 4	NCERT Solutions for Class 10 Maths Chapter 4
Chapter 5	NCERT Solutions for Class 10 Maths Chapter 5
Chapter 6	NCERT Solutions for Class 10 Maths Chapter 6
Chapter 7	NCERT Solutions for Class 10 Maths Chapter 7
Chapter 8	NCERT Solutions for Class 10 Maths Chapter 8
Chapter 9	NCERT Solutions for Class 10 Maths Chapter 9
Chapter 10	NCERT Solutions for Class 10 Maths Chapter 10
Chapter 11	NCERT Solutions for Class 10 Maths Chapter 11
Chapter 12	NCERT Solutions for Class 10 Maths Chapter 12
Chapter 13	NCERT Solutions for Class 10 Maths Chapter 13
Chapter 14	NCERT Solutions for Class 10 Maths Chapter 14

Q.1 Write first four terms of the AP, when the first terma and the common difference d are given as follows:(i) a=10, d=10 (ii) a=−2, d=0(iii) a=4, d=−3 (iv) a=−1, d=12(v) a=−1.25, d=−0.25

Ans.

$\begin{array}{l} (i) \\ We have, \\ First term = a = 10 \\ and \\ Common difference = d = 10 \\ Now, \\ Second term = a + d = 10 + 10 = 20, \\ Third term = a + 2 d = 10 + 2 \times 10 = 30, \\ Fourth term = a + 3 d = 10 + 3 \times 10 = 40, \\ Therefore, first four terms of the AP are: \\ 10, 20, 30, 40 \\ (ii) \\ We have, \\ First term = a = - 2 \\ and \\ Common difference = d = 0 \\ Now, \\ Second term = a + d = - 2 + 0 = - 2, \\ Third term = a + 2 d = - 2 + 2 \times 0 = - 2, \\ Fourth term = a + 3 d = - 2 + 3 \times 0 = - 2, \\ Therefore, first four terms of the AP are: \\ - 2, - 2, - 2, - 2 \\ (iii) \\ We have, \\ First term = a = 4 \\ and \\ Common difference = d = - 3 \\ Now, \\ Second term = a + d = 4 - 3 = 1, \\ Third term = a + 2 d = 4 + 2 \times (- 3) = - 2, \\ Fourth term = a + 3 d = 4 + 3 \times (- 3) = - 5, \\ Therefore, first four terms of the AP are: \\ 4, 1, - 2, - 5 \\ (iv) \\ We have, \\ First term = a = - 1 \\ and \\ Common difference = d = \frac{1}{2} \\ Now, \\ Second term = a + d = - 1 + \frac{1}{2} = - \frac{1}{2} \\ Third term = a + 2 d = - 1 + 2 \times \frac{1}{2} = 0 \\ Fourth term = a + 3 d = - 1 + 3 \times \frac{1}{2} = \frac{1}{2} \\ Therefore, first four terms of the AP are: \\ - 1, - \frac{1}{2}, 0, \frac{1}{2} \\ (v) \\ We have, \\ First term = a = - 1 .25 \\ and \\ Common difference = d = - 0 .25 \\ Now, \\ Second term = a + d = - 1 .25 - 0 .25 = - 1 .50, \\ Third term = a + 2 d = - 1 .25 + 2 \times (- 0 .25) = - 1 .75, \\ Fourth term = a + 3 d = - 1 .25 + 3 \times (- 0 .25) = - 2, \\ Therefore, first four terms of the AP are: \\ - 1 .25, - 1 .50, - 1 .75, - 2 \end{array}$

Q.2 For the following APs, write the first term and thecommon difference:(i) 3, 1, −1, −3, . . . (ii)−5, −1, 3, 7, . . .(iii) 13, 53, 93, 133,. . . (iv) 0.6, 1.7, 2.8, 3.9, . . .

Ans.

$\begin{array}{l} (i) \\ Following is the given AP. \\ 3, 1, - 1, - 3, . . . \\ Here, first term = 3 \\ and \\ common difference = difference between a term and \\ its preceding term \\ = 1 - 3 = - 2 \\ (ii) \\ Following is the given AP. \\ - 5, - 1, 3, 7, . . . \\ Here, first term = - 5 \\ and \\ common difference = difference between a term and \\ its preceding term \\ = - 1 - 5 = - 6 \\ (iii) \\ Following is the given AP. \\ \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{9}, . . . \\ Here, first term = \frac{1}{3} \\ and \\ common difference = difference between a term and \\ its preceding term \\ = \frac{5}{3} - \frac{1}{3} = \frac{4}{3} \\ (iv) \\ Following is the given AP. \\ 0.6, 1.7, 2.8, 3.9, . . . \\ Here, first term = 0 .6 \\ and \\ common difference = difference between a term and \\ its preceding term \\ = 1 .7 - 0 .6 = 1.1 \end{array}$

Q.3 Which of the following are APs? If they form an AP,find the common difference d and write three moreterms.(i) 2, 4, 8, 16, . . .(ii) 2, 52, 3, 72, . . .(iii) −1.2, −3.2, −5.2, −7.2, . . .(iv) −10, −6,−2, 2, . . .(v) 3, 3+2, 3+22, 3+32, . . .(vi) 0.2, 0.22, 0.222, 0.2222, . . .(vii) 0, −4, −8, −12, . . .(viii) −12, −12, −12, −12, . . .(ix) 1, 3, 9, 27, ...(x) a, 2a, 3a, 4a, ...(xi) a, a2, a3, a4, ...(xii) 2, 8, 18, 32, ...(xiii) 3, 6, 9, 12, ...(xiv) 12, 32, 52, 72, ...(xv) 12, 52, 72, 73, ...

Ans.

$\begin{array}{l} (i) 2, 4, 8, 16, . . . \\ Here, 4 - 2 = 2 \neq 8 - 4 \\ Therefore, the given list of numbers is not an AP. \\ (ii) 2, \frac{5}{2}, 3, \frac{7}{2}, . . . \\ We have, \\ a_{2} - a_{1} = \frac{5}{2} - 2 = \frac{1}{2} \\ a_{3} - a_{2} = 3 - \frac{5}{2} = \frac{1}{2} \\ a_{4} - a_{3} = \frac{7}{2} - 3 = \frac{1}{2} \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = \frac{1}{2} . \\ The next three terms are: \frac{7}{2} + \frac{1}{2} = 4, 4 + \frac{1}{2} = \frac{9}{2} and \frac{9}{2} + \frac{1}{2} = 5 . \\ (iii) - 1.2, - 3.2, - 5.2, - 7.2, . . . \\ We have, \\ a_{2} - a_{1} = - 3.2 - (- 1.2) = - 3.2 + 1.2 = - 2 \\ a_{3} - a_{2} = - 5.2 - (- 3.2) = - 5.2 + 3.2 = - 2 \\ a_{4} - a_{3} = - 7.2 - (- 5.2) = - 7.2 + 5.2 = - 2 \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = - 2 . \\ The next three terms are: - 7.2 - 2 = - 9.2, - 9.2 - 2 = - 11.2 \\ and - 11.2 - 2 = - 13.2 . \\ (iv) - 10, - 6, - 2, 2, . . . \\ We have, \\ a_{2} - a_{1} = - 6 - (- 10) = - 6 + 10 = 4 \\ a_{3} - a_{2} = - 2 - (- 6) = - 2 + 6 = 4 \\ a_{4} - a_{3} = 2 - (- 2) = 2 + 2 = 4 \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = 4 . \\ The next three terms are: 2 + 4 = 6, 6 + 4 = 10 and 10 + 4 = 14 . \\ (v) 3, 3 + \sqrt{2}, 3 + 2 \sqrt{2}, 3 + 3 \sqrt{2}, . . . \\ We have, \\ a_{2} - a_{1} = 3 + \sqrt{2} - 3 = \sqrt{2} \\ a_{3} - a_{2} = 3 + 2 \sqrt{2} - (3 + \sqrt{2}) = \sqrt{2} \\ a_{4} - a_{3} = 3 + 3 \sqrt{2} - (3 + 2 \sqrt{2}) = \sqrt{2} \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = \sqrt{2} . \\ The next three terms are: 3 + 3 \sqrt{2} + \sqrt{2} = 3 + 4 \sqrt{2}, \\ 3 + 4 \sqrt{2} + \sqrt{2} = 3 + 5 \sqrt{2} and 3 + 5 \sqrt{2} + \sqrt{2} = 3 + 6 \sqrt{2} . \\ (vi) 0 .2, 0 .22, 0 .222, 0 .2222, . . . \\ We have, \\ a_{2} - a_{1} = 0 .22 - 0 .2 = 0 .02 \\ a_{3} - a_{2} = 0 .222 - 0 .22 = 0 .002 \\ {So, a}_{2} - a_{1} \neq a_{3} - a_{2} \\ Therefore, the given list of numbers is not an AP. \\ (vii) 0, - 4, - 8, - 12, . . . \\ We have, \\ a_{2} - a_{1} = - 4 - 0 = - 4 \\ a_{3} - a_{2} = - 8 - (- 4) = - 8 + 4 = - 4 \\ a_{4} - a_{3} = - 12 - (- 8) = - 12 + 8 = - 4 \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = - 4 . \\ The next three terms are: - 12 - 4 = - 16, - 16 - 4 = - 20 \\ and - 20 - 4 = - 24 . \\ (viii) - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, . . . \\ {Obviously, a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = 0 . \\ The next three terms are: - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2} . \\ (ix) 1, 3, 9, 27, . .. \\ We have, \\ a_{2} - a_{1} = 3 - 1 = 2 \\ a_{3} - a_{2} = 9 - 3 = 6 \\ {So, a}_{2} - a_{1} \neq a_{3} - a_{2} \\ Therefore, the given list of numbers is not an AP. \\ (x) a, 2a, 3a, 4a, . .. \\ We have, \\ a_{2} - a_{1} = 2 a - a = a \\ a_{3} - a_{2} = 3a - 2 a = a \\ a_{4} - a_{3} = 4a - 3a = a \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = a . \\ The next three terms are: 4 a + a = 5 a, 5 a + a = 6 a \\ and 6a + a = 7 a . \\ {(xi) a, a}^{2}, a^{3}, a^{4}, . .. \\ {Obviously, a}_{k + 1} - a_{k} is not the same every time. \\ Therefore, the given list of numbers is not an AP. \\ (xii) \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, . .. \\ We have, \\ a_{2} - a_{1} = \sqrt{8} - \sqrt{2} = 2 \sqrt{2} - \sqrt{2} = \sqrt{2} \\ a_{3} - a_{2} = \sqrt{18} - \sqrt{8} = 3 \sqrt{2} - 2 \sqrt{2} = \sqrt{2} \\ a_{4} - a_{3} = \sqrt{32} - \sqrt{18} = 4 \sqrt{2} - 3 \sqrt{2} = \sqrt{2} \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = \sqrt{2} . \\ The next three terms are: \sqrt{32} + \sqrt{2} = 4 \sqrt{2} + \sqrt{2} = 5 \sqrt{2} = \sqrt{50}, \\ \sqrt{50} + \sqrt{2} = 5 \sqrt{2} + \sqrt{2} = 6 \sqrt{2} = \sqrt{72} \\ and \sqrt{72} + \sqrt{2} = 6 \sqrt{2} + \sqrt{2} = 7 \sqrt{2} = \sqrt{98} . \\ (xiii) \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, . .. \\ We have, \\ a_{2} - a_{1} = \sqrt{6} - \sqrt{3} = \sqrt{2} \times \sqrt{3} - \sqrt{3} = \sqrt{3} (\sqrt{2} - 1) \\ a_{3} - a_{2} = \sqrt{9} - \sqrt{6} = \sqrt{3} \times \sqrt{3} - \sqrt{2} \times \sqrt{3} = \sqrt{3} (\sqrt{3} - \sqrt{2}) \\ {So, a}_{2} - a_{1} \neq a_{3} - a_{2} \\ Therefore, the given list of numbers is not an AP. \\ {(xiv) 1}^{2} {, 3}^{2}, 5^{2}, 7^{2}, . .. \\ We have, \\ a_{2} - a_{1} = 3^{2} - 1^{2} = 9 - 1 = 8 \\ a_{3} - a_{2} = 5^{2} - 3^{2} = 25 - 9 = 16 \\ {So, a}_{2} - a_{1} \neq a_{3} - a_{2} \\ Therefore, the given list of numbers is not an AP. \\ {(xv) 1}^{2} {, 5}^{2}, 7^{2}, 73, . .. \\ We have, \\ a_{2} - a_{1} = 5^{2} - 1^{2} = 25 - 1 = 24 \\ a_{3} - a_{2} = 7^{2} - 5^{2} = 49 - 25 = 24 \\ a_{4} - a_{3} = 73 - 7^{2} = 73 - 49 = 24 \\ {i.e., a}_{k + 1} - a_{k} is the same every time. \\ Therefore, the given list of numbers is an AP with the common \\ difference d = 24 . \\ The next three terms are: 73 + 24 = 97, 97 + 24 = 121 \\ and 121 + 24 = 145 . \end{array}$

Q.4 Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP:

Ans.

$\begin{array}{l} (i) \\ We have \\ a = 7, d = 3, n = 8 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow a_{n} = 7 + (8 - 1) 3 = 28 \\ (ii) \\ We have \\ a = - 18, n = 10, a_{n} = 0 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow 0 = - 18 + (10 - 1) d \\ \Rightarrow d = \frac{18}{9} = 2 \\ (iii) \\ We have \\ d = - 3, n = 18, a_{n} = - 5 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow - 5 = a + (18 - 1) (- 3) \\ \Rightarrow a = 51 - 5 = 46 \\ (iv) \\ We have \\ a = - 18 .9, d = 2 .5, a_{n} = 3 .6 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow 3 .6 = - 18 .9 + (n - 1) (2 .5) \\ \Rightarrow (n - 1) (2 .5) = 3 .6 + 18 .9 = 22 .5 \\ \Rightarrow n = \frac{22 .5}{2 .5} + 1 = 10 \\ (v) \\ We have \\ a = 3 .5, d = 0, n = 105 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow a_{n} = 3 .5 + (105 - 1) \times 0 = 3 .5 \end{array}$

Q.5 Choose the correct choice in the following and justify:(i) 30th term of the AP: 10, 7, 4, . . . , is (A) 97 (B) 77 (C) −77 (D) −87(ii) 11th term of the AP: −3, −12, 2, . . ., is (A) 28 (B) 22 (C) −38 (D) −4812

Ans.

$\begin{array}{l} (i) \\ The given AP is: \\ 10, 7, 4, . . . \\ Here, a = 10, d = 7 - 10 = - 3, n = 30 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow a_{30} = 10 + (30 - 1) (- 3) = 10 - 87 = - 77 \\ Hence, the correct answer is (C). \\ (ii) \\ The given AP is: \\ - 3, \frac{- 1}{2}, 2, . . . \\ Here, a = - 3, d = 2 - \frac{- 1}{2} = \frac{5}{2}, n = 11 \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow a_{11} = - 3 + (11 - 1) \times \frac{5}{2} = - 3 + 25 = 22 \\ Hence, the correct answer is (B). \end{array}$

Q.6 In the following APs, find the missing terms in the boxes:i 2, , 26ii , 13, , 3iii 5, , , 912iv−4, , , , , 6v , 38, , , , −22

Ans.

$\begin{array}{l} (i) \\ The given AP is : \\ 2,, 26 \\ Let the missing term is ‘ t ‘ . Then, we have \\ t - 2 = 26 - t \\ \Rightarrow 2 t = 26 + 2 = 28 \\ \Rightarrow t = \frac{28}{2} = 14 \\ (ii) \\ The given AP is : \\ , 13,, 3 \\ Let the common difference is ‘ d ‘ . Then, we have \\ a_{4} - a_{2} = 2 d = 3 - 13 \\ \Rightarrow 2 d = - 10 \\ \Rightarrow d = \frac{- 10}{2} = - 5 \\ Therefore, \\ first term = a_{1} = a_{2} - d = 13 - (- 5) = 13 + 5 = 18 \\ third term = a_{3} = a_{4} - d = 3 - (- 5) = 3 + 5 = 8 \\ Hence, the given APis : \\ 18, 13, 8, 3 \\ (iii) \\ The given AP is : \\ 5,,, 9 \frac{1}{2} \\ Let the common difference is ‘ d ‘ . Then, we have \\ a_{2} = a_{1} + d, \\ a_{3} = a_{2} + d = a_{1} + d + d = a_{1} + 2 d, \\ a_{4} = a_{3} + d = a_{1} + 2 d + d = a_{1} + 3 d \\ \Rightarrow \frac{19}{2} = 5 + 3 d \\ \Rightarrow 3 d = \frac{19}{2} - 5 = \frac{9}{2} \\ \Rightarrow d = \frac{9}{2} \times \frac{1}{3} = \frac{3}{2} \\ Therefore, missing terms of the given AP are : \\ a_{2} = a_{1} + d = 5 + \frac{3}{2} = \frac{13}{2}, \\ a_{3} = a_{2} + d = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8 \\ Hence, the given AP is : \\ 5, \frac{13}{2}, 8, 9 \frac{1}{2} \\ (iv) \\ The given AP is : \\ - 4,,,,, 6 \\ Let the common difference is ‘ d ‘ . Then, we have \\ a_{6} = a_{1} + 5 d \\ \Rightarrow 6 = - 4 + 5 d \\ \Rightarrow 5 d = 6 + 4 = 10 \\ \Rightarrow d = \frac{10}{5} = 2 \\ Therefore, missing terms of the given AP are : \\ a_{2} = a_{1} + d = - 4 + 2 = - 2 \\ a_{3} = a_{1} + 2 d = - 4 + 2 \times 2 = 0, \\ a_{4} = a_{1} + 3 d = - 4 + 3 \times 2 = 2, \\ a_{5} = a_{1} + 4 d = - 4 + 4 \times 2 = 4 \\ Hence, the given AP is : \\ - 4, - 2, 0, 2, 4, 6 \\ (v) \\ The given AP is : \\ , 38,,,, - 22 \\ Let the common difference is ‘ d ‘ . Then, we have \\ a_{6} = a_{1} + 5 d \\ \Rightarrow a_{6} = a_{1} + d + 4 d = a_{2} + 4 d \\ \Rightarrow - 22 = 38 + 4 d \\ \Rightarrow 4 d = - 22 - 38 = - 60 \\ \Rightarrow d = \frac{- 60}{4} = - 15 \\ Therefore, missing terms of the given AP are : \\ a_{1} = a_{2} + d = 38 - (- 15) = 38 + 15 = 53, \\ a_{3} = a_{1} + 2 d = 53 + 2 \times (- 15) = 23, \\ a_{4} = a_{1} + 3 d = 53 + 3 \times (- 15) = 8 \\ a_{5} = a_{1} + 4 d = 53 + 4 \times (- 15) = - 7 \\ Hence, the given AP is : \\ 53, 38, 23, 8, - 7, - 22 \end{array}$

Q.7 Which term of the AP: 3, 8, 13, 18, . . . , is 78?

Ans.

$\begin{array}{l} The given AP is: \\ 3, 8, 13, 18, . . . \\ {Let n}^{th} term of the given AP is 78. \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow 78 = 3 + (n - 1) 5 \\ \Rightarrow (n - 1) 5 = 78 - 3 = 75 \\ \Rightarrow n - 1 = \frac{75}{5} = 15 \\ \Rightarrow n = 15 + 1 = 16 \\ {Therefore, 16}^{th} term of the given AP is 78. \end{array}$

Q.8 Find the number of terms in each of the following APs:(i) 7, 13, 19, . . . , 205 (ii) 18, 1512, 13, . . . , −47

Ans.

$\begin{array}{l} (i) 7, 13, 19, . . . , 205 \\ Let the given AP has n terms. \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow 205 = 7 + (n - 1) 6 \\ \Rightarrow (n - 1) 6 = 205 - 7 = 198 \\ \Rightarrow n - 1 = \frac{198}{6} = 33 \\ \Rightarrow n = 33 + 1 = 34 \\ (ii) 18, 15 \frac{1}{2}, 13, . . . , - 47 \\ Let the given AP has n terms. \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow - 47 = 18 + (n - 1) (\frac{31}{2} - 18) = 18 + (n - 1) (\frac{- 5}{2}) \\ \Rightarrow (n - 1) (\frac{- 5}{2}) = - 47 - 18 = - 65 \\ \Rightarrow n - 1 = - 65 \times \frac{- 2}{5} = 26 \\ \Rightarrow n = 26 + 1 = 27 \end{array}$

Q.9 Check whether –150 is a term of the AP: 11, 8, 5, 2, . . .

Ans.

$\begin{array}{l} The given AP is 11, 8, 5, 2, . . . \\ {Let n}^{th} term of the given AP is - 150. \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow - 150 = 11 + (n - 1) (8 - 11) = 11 + (n - 1) (- 3) \\ \Rightarrow (n - 1) (- 3) = - 150 - 11 = - 161 \\ \Rightarrow n - 1 = - 161 \times \frac{- 1}{3} = 53 \frac{2}{3} \\ \Rightarrow n = 53 \frac{2}{3} + 1 = 54 \frac{2}{3} \\ But 54 \frac{2}{3} is not a natural number. Therefore, - 150 is not \\ a term of the given AP. \end{array}$

Q.10 Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Ans.

$\begin{array}{l} We have \\ a_{11} = 38 and a_{16} = 73 \\ {Or a}_{11} = a_{1} + 10 d = 38 . .. (1) \\ and a_{16} = a_{1} + 15 d = 73 . .. (2) \\ Subtracting (1) from (2), we get \\ 5 d = 35 \\ \Rightarrow d = 7 \\ On putting this value of d in (1), we get \\ a_{1} + 10 d = 38 \\ \Rightarrow a_{1} + 10 \times 7 = 38 \\ \Rightarrow a_{1} = 38 - 70 = - 32 \\ Now, \\ a_{31} = a_{1} + 30 d = - 32 + 30 \times 7 = - 32 + 210 = 178 \end{array}$

Q.11 An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29^th term.

Ans.

$\begin{array}{l} We have \\ a_{3} = 12 and a_{50} = 106 \\ {Or a}_{3} = a_{1} + 2 d = 12 . .. (1) \\ and a_{50} = a_{1} + 49 d = 106 . .. (2) \\ Subtracting (1) from (2), we get \\ 47 d = 94 \\ \Rightarrow d = \frac{94}{47} = 2 \\ On putting this value of d in (1), we get \\ a_{1} + 2 \times 2 = 12 \\ \Rightarrow a_{1} = 12 - 4 = 8 \\ Now, \\ a_{29} = a_{1} + 28 d = 8 + 28 \times 2 = 8 + 56 = 64 \end{array}$

Q.12 If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?

Ans.

$\begin{array}{l} We have \\ a_{3} = 4 and a_{9} = - 8 \\ {Or a}_{3} = a_{1} + 2 d = 4 . .. (1) \\ and a_{9} = a_{1} + 8 d = - 8 . .. (2) \\ Subtracting (1) from (2), we get \\ 6 d = - 12 \\ \Rightarrow d = \frac{- 12}{6} = - 2 \\ On putting this value of d in (1), we get \\ a_{1} + 2 \times (- 2) = 4 \\ \Rightarrow a_{1} = 4 + 4 = 8 \\ {Let n}^{th} term of the given AP is zero. Then, we have \\ a_{n} = a_{1} + (n - 1) d \\ \Rightarrow 0 = 8 + (n - 1) (- 2) \\ \Rightarrow n - 1 = \frac{- 8}{- 2} = 4 \\ \Rightarrow n = 4 + 1 = 5 \\ Hence, 5th term of the given AP is zero. \end{array}$

Q.13 The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Ans.

$\begin{array}{l} We have \\ a_{17} = a_{10} + 7 \\ \Rightarrow a_{17} - a_{10} = 7 \\ \Rightarrow a_{1} + 16 d - (a_{1} + 9 d) = 7 [a_{n} = a + (n - 1) d] \\ \Rightarrow 7 d = 7 \\ \Rightarrow d = 1 \\ Therefore, the common difference is 1. \end{array}$

Q.14 Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Ans.

$\begin{array}{l} Let n^{th} {term be 132 more than 54}^{th} term of the given AP. \\ We have \\ a_{n} = a_{54} + 132 \\ \Rightarrow a_{n} - a_{54} = 132 \\ \Rightarrow a_{1} + (n - 1) d - (a_{1} + 53 d) = 132 [a_{n} = a + (n - 1) d] \\ \Rightarrow (n - 54) d = 132 \\ \Rightarrow n - 54 = \frac{132}{d} = \frac{132}{12} [The given AP is: 3, 15, 27, 39, . ..] \\ \Rightarrow n = 11 + 54 = 65 \\ Therefore, 65^{th} {term is 132 more than 54}^{th} term of the given AP. \end{array}$

Q.15 Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Ans.

$\begin{array}{l} It is given that two APs have the same common difference. \\ Let a_{n} and b_{n} represent the terms of the first and second \\ AP respectively for all natural numbers n . \\ We have \\ a_{100} - b_{100} = 100 \\ \Rightarrow a_{1} + 99 d - b_{1} - 99 d = 100 [a_{n} = a + (n - 1) d] \\ \Rightarrow a_{1} - b_{1} = 100 \\ Now, \\ a_{1000} - b_{1000} = a_{1} + 999 d - b_{1} - 999 d = a_{1} - b_{1} = 100 \\ Therefore, the difference between 1000th terms of the two \\ APs is 100. \end{array}$

Q.16 How many three-digit numbers are divisible by 7?

Ans.

$\begin{array}{l} The smallest three-digit number divisible by 7 is 105. \\ The largest three-digit number divisible by 7 is 994. \\ Three-digit numbers divisible by 7 are: \\ 105, 112, 119, . ..,994 \\ These multiples of 7 form an AP with common difference 7, \\ first term 105 and last term 994. Thus, we have \\ a_{1} = 105, a_{n} = 994 and d = 7 . \\ We know that \\ a_{n} = a_{1} + (n - 1) d \\ \Rightarrow 994 = 105 + (n - 1) 7 \\ \Rightarrow (n - 1) 7 = 994 - 105 = 889 \\ \Rightarrow n - 1 = \frac{889}{7} = 127 \\ \Rightarrow n = 127 + 1 = 128 \\ Therefore, 128 three-digit numbers are divisible by 7. \end{array}$

Q.17 How many multiples of 4 lie between 10 and 250?

Ans.

$\begin{array}{l} Multiples of 4 between 10 and 250 are: \\ 12, 16, 20, . ..,248 \\ These multiples of 4 form an AP with common difference 4, \\ first term 12 and last term 248. Thus, we have \\ a_{1} = 12, a_{n} = 248 and d = 4 . \\ We know that \\ a_{n} = a_{1} + (n - 1) d \\ \Rightarrow 248 = 12 + (n - 1) 4 \\ \Rightarrow (n - 1) 4 = 248 - 12 = 236 \\ \Rightarrow n - 1 = \frac{236}{4} = 59 \\ \Rightarrow n = 59 + 1 = 60 \\ Therefore, 60 multiples of 4 lie between 10 and 250 . \end{array}$

Q.18 For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Ans.

$\begin{array}{l} Let n th term of the given APs are equal. \\ Let a_{n} and b_{n} represent the terms of the first and second \\ AP respectively for all natural numbers n . Also, let d_{1} and d_{2} are \\ common diffrences of first and second AP respectively. \\ The given two APs are: \\ 63, 65, 67, . . . and 3, 10, 17, . . . \\ We have \\ a_{n} = b_{n} \\ \Rightarrow a_{1} + (n - 1) d_{1} = b_{1} + (n - 1) d_{2} [a_{n} = a + (n - 1) d] \\ \Rightarrow 63 + (n - 1) 2 = 3 + (n - 1) 7 \\ \Rightarrow (n - 1) 2 - (n - 1) 7 = 3 - 63 \\ \Rightarrow - 5 (n - 1) = - 60 \\ \Rightarrow n - 1 = \frac{- 60}{- 5} = 12 \\ \Rightarrow n = 12 + 1 = 13 \\ Therefore, 13th terms of the given APs are equal. \end{array}$

Q.19 Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Ans.

$\begin{array}{l} We have \\ a_{3} = 16 and a_{7} = a_{5} + 12 \\ \Rightarrow a_{1} + 2 d = 16 and a_{7} - a_{5} = 12 [a_{n} = a + (n - 1) d] \\ \Rightarrow a_{1} + 2 d = 16 and a_{1} + 6 d - a_{1} - 4 d = 12 \\ \Rightarrow a_{1} + 2 d = 16 and 2 d = 12 \\ \Rightarrow a_{1} + 2 d = 16 and d = 6 \\ \Rightarrow a_{1} + 2 \times 6 = 16 and d = 6 \\ \Rightarrow a_{1} = 4 and d = 6 \\ Therefore, the required AP is: 4, 10, 16, 22, . .. \end{array}$

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

Ans.

$\begin{array}{l} Here, a = 3, d = 8 - 3 = 5, l = 253 where l = a + (n - 1) d \\ To find the 20th term from the last term, we will find the \\ total number of terms in the AP. \\ So, 253 = 3 + (n - 1) 5 \\ i.e., 253 - 3 = (n - 1) 5 \\ i.e., n - 1 = \frac{250}{5} = 50 \\ or n = 50 + 1 = 51 \\ So, there are 51 terms in the given AP. \\ The 20th term from the last term will be the 32nd term(51-20+1) from the beginning. \\ {So, a}_{32} = 3 + 31 \times 5 = 3 + 155 = 158 \\ i.e., the 20th term from the last term is 158. \end{array}$

Q.21 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.

Ans.

$\begin{array}{l} Here, \\ a_{4} + a_{8} = 24 \\ \Rightarrow a + 3 d + a + 7 d = 24 \\ \Rightarrow 2 a + 10 d = 24 \\ \Rightarrow a + 5 d = 12 . .. (1) \\ Also, \\ a_{6} + a_{10} = 44 \\ \Rightarrow a + 5 d + a + 9 d = 44 \\ \Rightarrow 2 a + 14 d = 44 \\ \Rightarrow a + 7 d = 22 . .. (2) \\ We subtract (2) from (1) and get \\ 2 d = 10 \\ \Rightarrow d = 5 \\ On putting this value of d in (1), we get \\ a + 5 \times 5 = 12 \\ \Rightarrow a = 12 - 25 = - 13 \\ First three terms of the AP are a, a + d and a + 2 d i.e., \\ - 13, - 13 + 5 and - 13 + 2 \times 5 i.e., - 1 3, - 8 and - 3. \end{array}$

Q.22 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?

Ans.

$\begin{array}{l} Subba Rao’s annual salary in 1995 was ₹ 5000. \\ He received an increment of ₹ 200 each year. \\ So, Subba Rao’s annual salaries forms an AP with \\ a=5000 and d=200. \\ {Let 7000 is n}^{th} term of this AP. \\ We know that \\ a_{n} =a+(n-1)d \\ \Rightarrow 7000=5000+(n-1)200 \\ \Rightarrow (n-1)200=7000-5000 \\ \Rightarrow n-1= \frac{2000}{200} =10 \\ \Rightarrow n=10+1=11 \\ {Therefore, in 11}^{th} year, Subba Rao’s salary reach ₹ 7000. \end{array}$

Q.23 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.

Ans.

$\begin{array}{l} ₹ \\ ₹ . . \\ So, ‘ s forms an AP with \\ a = 5 and d = . . \\ {It is given that in n}^{th} week her weekly savings become \\ ‘ . . \\ ∴ a_{n} = . \\ We know that \\ a_{n} = a + (n - 1) d \\ \Rightarrow . = 5 + (n - 1) \times . \\ \Rightarrow (n - 1) \times . = . - 5 \\ \Rightarrow n - 1 = \frac{15 .75}{.} = 9 \\ \Rightarrow n = 9 + 1 = 10 \end{array}$

Q.24

$\begin{array}{l} Find the sum of the following APs: \\ (i) 2, 7, 12, . . ., to 10 terms. \\ (ii) - 37, - 33, - 29, . . ., to 12 terms. \\ (iii) 0.6, 1.7, 2.8, . . ., to 100 terms. \\ (iv) \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, . . ., to 11 terms. \end{array}$

Ans.

$\begin{array}{l} (i) \\ Here, a = 2, d = a_{2} - a_{1} = 7 - 2 = 5, n = 10 \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} [2 a + (n - 1) d] \\ So, S = \frac{10}{2} [2 \times 2 + (10 - 1) 5] = 5 [4 + 45] \\ or S = 245 \\ So, the sum of the first 10 terms of the given AP is 245. \\ (ii) \\ Here, a = - 37, d = a_{2} - a_{1} = - 33 - (- 37) = 4, n = 12 \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} [2 a + (n - 1) d] \\ So, S = \frac{12}{2} [2 \times (- 37) + (12 - 1) 4] = 6 [- 74 + 44] \\ or S = - 180 \\ So, the sum of the first 12 terms of the given AP is - 180 . \\ (iii) \\ Here, a = 0.6, d = a_{2} - a_{1} = 1.7 - 0.6 = 1 .1, n = 100 \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} [2 a + (n - 1) d] \\ So, S = \frac{100}{2} [2 \times 0.6 + (100 - 1) \times 1 .1] = 50 [1 .2 + 108 .9] \\ or S = 5505 \\ So, the sum of the first 100 terms of the given AP is 5505 . \\ (iv) \\ Here, a = \frac{1}{15}, d = a_{2} - a_{1} = \frac{1}{12} - \frac{1}{15} = \frac{1}{60}, n = 11 \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} [2 a + (n - 1) d] \\ So, S = \frac{11}{2} [2 \times \frac{1}{15} + (11 - 1) \times \frac{1}{60}] = \frac{11}{2} [\frac{2}{15} + \frac{1}{6}] \\ or S = \frac{33}{20} \\ So, the sum of the first 11 terms of the given AP is \frac{33}{20} . \end{array}$

Q.25

$\begin{array}{l} Find the sums given below : \\ (i) 7 + 10 \frac{1}{2} + 14 + . . . + 84 \\ (ii) 34 + 32 + 30 + . . . + 10 \\ (iii) - 5 + (- 8) + (- 11) + . . . + (- 230) \end{array}$

Ans.

$\begin{array}{l} (i) 7 + 10 \frac{1}{2} + 14 + . . . + 84 \\ Here, a = 7, d = a_{2} - a_{1} = 10 \frac{1}{2} - 7 = 3 \frac{1}{2} = \frac{7}{2}, l = 84 \\ We know that \\ l = a + (n - 1) d \\ So, 84 = 7 + (n - 1) \frac{7}{2} \\ or 84 - 7 = (n - 1) \frac{7}{2} \\ or 77 \times \frac{2}{7} + 1 = n \\ or n = 23 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + l) \\ So, S_{23} = \frac{23}{2} (7 + 84) = \frac{23}{2} \times 91 \\ or S_{23} = 1046 .5 \\ So, the requird sum is 1046 .5. \\ (ii) 34 + 32 + 30 + . . . + 10 \\ Here, a = 34, d = a_{2} - a_{1} = 32 - 34 = - 2, l = 10 \\ We know that \\ l = a + (n - 1) d \\ So, 10 = 34 + (n - 1) (- 2) \\ or 10 - 34 = (n - 1) (- 2) \\ or \frac{- 24}{- 2} = n - 1 \\ or n = 13 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + l) \\ So, S_{13} = \frac{13}{2} (34 + 10) = \frac{13}{2} \times 44 \\ or S_{13} = 286 \\ So, the requird sum is 286 . \\ (iii) - 5 + (- 8) + (- 11) + . . . + (- 230) \\ Here, a = - 5, d = a_{2} - a_{1} = - 8 - (- 5) = - 3, l = - 230 \\ We know that \\ l = a + (n - 1) d \\ So, - 230 = - 5 + (n - 1) (- 3) \\ or - 230 + 5 = (n - 1) (- 3) \\ or \frac{- 225}{- 3} + 1 = n \\ or n = 76 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + l) \\ So, S_{75} = \frac{76}{2} (- 5 - 230) = 38 \times (- 235) \\ or S_{75} = - 8930 \\ So, the requird sum is - 8930 . \end{array}$

Q.26

$\begin{array}{l} In an AP: \\ (i) given a = 5, d = 3, a_{n} = 50, find n and S_{n} . \\ (ii) given a = 7, a_{13} = 35, find d and S_{13} . \\ (iii) given a_{12} = 37, d = 3, find a and S_{12} . \\ (iv) given a_{3} = 15, S_{10} = 125, find d and a_{10} . \\ (v) given d = 5, S_{9} = 75, find a and a_{9} . \\ (vi) given a = 2, d = 8, S_{n} = 90, find n and a_{n} . \\ (vii) given a = 8, a_{n} = 62, S_{n} = 210, find n and d . \\ (viii) given a_{n} = 4, d = 2, S_{n} = - 14, find n and a . \\ (ix) given a = 3, n = 8, S = 192, find d . \\ (x) given l = 28, S = 144, and there are total 9 terms. \\ Find a . \end{array}$

Ans.

$\begin{array}{l} (i) \\ Here, a = 5, d = 3, a_{n} = 50 \\ We have to find n and S_{n} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, 50 = 5 + (n - 1) 3 \\ or 50 - 5 = (n - 1) 3 \\ or \frac{45}{3} + 1 = n \\ or n = 16 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, S_{16} = \frac{16}{2} (5 + 50) = 8 \times 55 \\ or S_{16} = 440 \\ (ii) \\ Here, a = 7, a_{13} = 35 \\ We have to find d and S_{13} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{13} = 7 + (13 - 1) d \\ or 35 - 7 = 12 d \\ or \frac{28}{12} = d \\ or d = \frac{7}{3} \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, S_{13} = \frac{13}{2} (7 + 35) = \frac{13}{2} \times 42 \\ or S_{13} = 273 \\ (iii) \\ Here, a_{12} = 37, d = 3 \\ We have to find a and S_{12} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{12} = a + (12 - 1) 3 \\ or 37 = a + 33 \\ or a = 37 - 33 = 4 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, S_{12} = \frac{12}{2} (a + a_{12}) = 6 (4 + 37) \\ or S_{12} = 246 \\ (iv) \\ Here, a_{3} = 15, S_{10} = 125 \\ We have to find d and a_{10} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{3} = a + (3 - 1) d \\ or a + 2 d = 15 . .. (1) \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{10} = \frac{10}{2} [2 a + (10 - 1) d] = 5 [2 a + 9 d] \\ or 125 = 5 [2 a + 9 d] \\ or 2 a + 9 d = 25 \\ or 2 (15 - 2 d) + 9 d = 25 [From (1), a = 15 - 2 d] \\ or 5 d = 25 - 30 = - 5 \\ or d = - 1 \\ putting this value of d in (1), we get a = 17 . \\ Now, \\ a_{10} = a + (10 - 1) d = 17 + 9 \times (- 1) = 8 \\ (v) \\ Here, d = 5, S_{9} = 75 \\ We have to find a and a_{9} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{9} = a + (9 - 1) 5 \\ or a_{9} - a = 40 . .. (1) \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{9} = \frac{9}{2} [2 a + (9 - 1) 5] = \frac{9}{2} [2 a + 40] \\ or 75 = 9 [a + 20] \\ or a = \frac{75}{9} - 20 = \frac{75 - 180}{9} = \frac{- 105}{9} = \frac{- 35}{3} \\ or a = \frac{- 35}{3} \\ On putting this value of a in (1), we get \\ a_{9} - \frac{- 35}{3} = 40 \\ \Rightarrow a_{9} = 40 - \frac{35}{3} = \frac{120 - 35}{3} = \frac{85}{3} \\ (vi) \\ Here, a = 2, d = 8, S_{n} = 90 \\ We have to find n and a_{n} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{n} = 2 + (n - 1) 8 . .. (1) \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{n} = \frac{n}{2} [2 \times 2 + (n - 1) 8] \\ or 90 = 2 n + 4 n (n - 1) = 2 n + 4 n^{2} - 4 n \\ or 90 = 4 n^{2} - 2 n \\ or 2 n^{2} - n - 45 = 0 \\ or 2 n^{2} - 10 n + 9 n - 45 = 0 \\ or 2 n (n - 5) + 9 (n - 5) = 0 \\ or (2 n + 9) (n - 5) = 0 \\ or n = 5 \\ On putting this value of n in (1), we get \\ a_{5} = 2 + (5 - 1) 8 = 34 \\ (vii) \\ Here, a = 8, a_{n} = 62, S_{n} = 210 \\ We have to find n and d . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, 62 = 8 + (n - 1) d . .. (1) \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, 210 = \frac{n}{2} (8 + 62) \\ or 210 = 35 n \\ or n = \frac{210}{35} = 6 \\ On putting this value of n in (1), we get \\ 62 = 8 + (n - 1) d = 8 + (6 - 1) d = 8 + 5 d \\ or d = \frac{62 - 8}{5} = \frac{54}{5} \\ (viii) \\ Here, a_{n} = 4, d = 2, S_{n} = - 14 \\ We have to find n and a . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, 4 = a + 2 (n - 1) = a + 2 n - 2 \\ or a + 2 n = 6 \\ or a = 6 - 2 n . .. (1) \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, - 14 = \frac{n}{2} (a + 4) \\ or n (a + 4) = - 28 \\ or n (6 - 2 n + 4) = - 28 [From (1)] \\ or 10 n - 2 n^{2} + 28 = 0 \\ or n^{2} - 5 n - 14 = 0 \\ or n^{2} - 7 n + 2 n - 14 = 0 \\ or n (n - 7) + 2 (n - 7) = 0 \\ or (n - 7) (n + 2) = 0 \\ or n - 7 = 0 \\ or n = 7 [n is a natural number] \\ On putting this value of n in (1), we get \\ a = 6 - 2 n = 6 - 2 \times 7 = - 8 \\ (ix) \\ Here, a = 3, n = 8, S = 192 \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} [2 a + (n - 1) d] \\ So, 192 = \frac{8}{2} [2 \times 3 + (8 - 1) d] \\ or 192 = 4 [6 + 7 d] \\ or 6 + 7 d = \frac{192}{4} = 48 \\ or 7 d = 48 - 6 = 42 \\ or d = \frac{42}{7} = 6 \\ (x) \\ Here, l = 28, S = 144, n = 9 \\ We have to find a . \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} (a + l) \\ So, 144 = \frac{9}{2} (a + 28) \\ or a + 28 = 144 \times \frac{2}{9} = 16 \times 2 = 32 \\ or a = 32 - 28 = 4 \end{array}$

Q.27

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

Ans.

$\begin{array}{l} Here, a = 9, d = a_{2} - a_{1} = 17 - 9 = 8, S_{n} = 636 \\ We know that sum of first n terms of an AP is given by \\ S = \frac{n}{2} [2 a + (n - 1) d] \\ So, 636 = \frac{n}{2} [2 \times 9 + (n - 1) 8] \\ or 636 = \frac{n}{2} \times 2 [9 + 4 n - 4] = 4 n^{2} + 5 n \\ or 4 n^{2} + 5 n - 636 = 0 \\ or 4 n^{2} + 53 n - 48 n - 636 = 0 \\ or n (4 n + 53) - 12 (4 n + 53) = 0 \\ or (4 n + 53) (n - 12) = 0 \\ or n = 12 [n is a natural number] \\ Therefore, 12 terms of the given AP must be taken to give a \\ sum of 636. \end{array}$

Q.28 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.

Ans.

$\begin{array}{l} Here, a = 5, a_{n} = l = 45, S_{n} = 400 \\ We have to find n and d . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, 45 = 5 + (n - 1) d . .. (1) \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, 400 = \frac{n}{2} (5 + 45) \\ or 400 = 25 n \\ or n = \frac{400}{25} = 16 \\ On putting this value of n in (1), we get \\ 45 = 5 + (n - 1) d = 5 + (16 - 1) d = 5 + 15 d \\ or d = \frac{45 - 5}{15} = \frac{40}{15} = \frac{8}{3} \end{array}$

Q.29 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?

Ans.

$\begin{array}{l} Here, a = 17, a_{n} = l = 350, d = 9 \\ We have to find n and S_{n} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, 350 = 17 + (n - 1) 9 \\ or n - 1 = \frac{350 - 17}{9} = \frac{333}{9} = 37 \\ or n = 37 + 1 = 38 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, S_{38} = \frac{38}{2} (17 + 350) = 19 \times 367 \\ or S_{38} = 6973 \end{array}$

Q.30 Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Ans.

$\begin{array}{l} Here, a_{22} = 149, d = 7 \\ We have to find S_{22} . \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{22} = a + (22 - 1) d \\ or 149 = a + 21 \times 7 = a + 147 \\ or a = 149 - 147 = 2 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} (a + a_{n}) \\ So, S_{22} = \frac{22}{2} (a + a_{22}) = 11 (2 + 149) = 11 \times 151 \\ or S_{22} = 1661 \end{array}$

Q.31 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Ans.

$\begin{array}{l} Here, a_{2} = 14, a_{3} = 18 \\ Therefore, d = a_{3} - a_{2} = 18 - 14 = 4 \\ We know that \\ a_{n} = a + (n - 1) d \\ So, a_{2} = a + (2 - 1) d \\ or 14 = a + 1 \times 4 = a + 4 \\ or a = 14 - 4 = 10 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{51} = \frac{51}{2} [2 \times 10 + (51 - 1) 4] = \frac{51}{2} (20 + 200) = 51 \times 110 \\ or S_{51} = 5610 \end{array}$

Q.32 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.

Ans.

$\begin{array}{l} Here, S_{7} = 49, S_{17} = 289 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{7} = \frac{7}{2} [2 a + (7 - 1) d] \\ or 49 = \frac{7}{2} (2 a + 6 d) \\ or \frac{49}{7} = \frac{2}{2} (a + 3 d) \\ or a + 3 d = 7 . .. (1) \\ Also, \\ S_{17} = \frac{17}{2} [2 a + (17 - 1) d] \\ or 289 = \frac{17}{2} (2 a + 16 d) = \frac{17}{2} \times 2 (a + 8 d) \\ or a + 8 d = 17 \\ or 7 - 3 d + 8 d = 17 [From (1), a = 7 - 3 d] \\ or 5 d = 17 - 7 = 10 \\ or d = \frac{10}{5} = 2 \\ On putting this value of d in (1), we get \\ a = 7 - 3 d = 7 - 3 \times 2 = 1 \\ Therefore, sum of first n terms of the AP with a = 1 and d = 2 is \\ given by \\ S_{n} = \frac{n}{2} [2 \times 1 + (n - 1) 2] = n (1 + n - 1) = n^{2} \end{array}$

Q.33 Show that a₁, a₂, . . ., a_n, . . . form an AP where a_n is defined as below :
(i) a_n = 3 + 4n (ii) a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.

Ans.

$\begin{array}{l} (i) \\ Here, \\ a_{n} = 3 + 4 n \\ \Rightarrow a_{n + 1} = 3 + 4 (n + 1) = 3 + 4 n + 4 = 7 + 4 n \\ Now, \\ a_{n + 1} - a_{n} = 7 + 4 n - 3 - 4 n = 4 \\ i.e., difference between a term and its preceding term is always 4. \\ Therefore, a_{1}, a_{2}, . . ., a_{n}, . . . form an AP where a_{n} = 3 + 4 n . \\ Now, we have \\ d = a_{n + 1} - a_{n} = 4 \\ and \\ a = a_{1} = 3 + 4 \times 1 = 7 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{15} = \frac{15}{2} [2 \times 7 + (15 - 1) 4] \\ or S_{15} = \frac{15}{2} \times 2 [7 + 14 \times 2] = 15 \times 35 = 525 \\ (ii) \\ Here, \\ a_{n} = 9 - 5 n \\ \Rightarrow a_{n + 1} = 9 - 5 (n + 1) = 9 - 5 n - 5 = 4 - 5 n \\ Now, \\ a_{n + 1} - a_{n} = 4 - 5 n - (9 - 5 n) = 4 - 5 n - 9 + 5 n = - 5 \\ i.e., difference between a term and its preceding term is \\ always - 5. \\ Therefore, a_{1}, a_{2}, . . ., a_{n}, . . . form an AP where a_{n} = 9 - 5 n . \\ Now, we have \\ d = a_{n + 1} - a_{n} = - 5 \\ and \\ a = a_{1} = 9 - 5 \times 1 = 4 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{15} = \frac{15}{2} [2 \times 4 + (15 - 1) (- 5)] \\ or S_{15} = \frac{15}{2} \times 2 [4 - 5 \times 7] = - 31 \times 15 = - 465 \end{array}$

Q.34 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.

Ans.

$\begin{array}{l} Here, \\ S_{} = - n^{2} \\ \Rightarrow S_{1} = \times 1 - 1^{2} = 3 \\ i.e., a = S_{1} = 3 \\ S_{2} = \times 2 - 2^{2} = 4 \\ Now, \\ a_{2} = S_{2} - S_{1} = (\times 2 - 2^{2}) - 3 = (8 - 4) - 3 = 1 \\ Similarly, \\ a_{3} = S_{3} - S_{2} = (\times 3 - 3^{2}) - (\times 2 - 2^{2}) \\ = (12 - 9) - (8 - 4) \\ = 3 - 4 = - 1 \\ a_{10} = S_{10} - S_{9} = (\times 10 - 10^{2}) - (\times 9 - 9^{2}) \\ = (40 - 100) - (36 - 81) = - 60 - (- 45) \\ = - 60 + 45 = - 15 \\ a_{n} = S_{n} - S_{n - 1} = (\times n - n^{2}) - (\times (n - 1) - {(n - 1)}^{2}) \\ = n (4 - n) - (n - 1) (4 - n + 1) \\ = n (4 - n) - (n - 1) (5 - n) \\ = 4 n - n^{2} - 5 n + n^{2} + 5 - n \\ = - 2 n + 5 \\ = 5 - 2 n \end{array}$

Q.35 Find the sum of the first 40 positive integers divisible by 6.

Ans.

$\begin{array}{l} First 40 positive integers divisible by 6 forms an AP with \\ a = 6 and d = 6 . \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{40} = \frac{40}{2} [2 \times 6 + (40 - 1) 6] = 20 [12 + 39 \times 6] \\ or S_{40} = 20 [12 + 234] = 20 \times 246 = 4920 \end{array}$

Q.36 Find the sum of the first 15 multiples of 8.

Ans.

$\begin{array}{l} First 15 multiples of 8 forms an AP with \\ a = 8 and d = 8 . \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{15} = \frac{15}{2} [2 \times 8 + (15 - 1) 8] = \frac{15}{2} \times 8 [2 + 14] \\ or S_{15} = 60 \times 16 = 960 \end{array}$

Q.37 Find the sum of the odd numbers between 0 and 50.

Ans.

$\begin{array}{l} Odd numbers between 0 and 50 forms an AP with \\ a = 1 and d = 2 . \\ Also, last term of this AP is 49. \\ Therefore, \\ l = a_{n} = 49 \\ \Rightarrow a + (n - 1) d = 49 \\ \Rightarrow 1 + (n - 1) 2 = 49 \\ \Rightarrow n - 1 = \frac{49 - 1}{2} = 24 \\ \Rightarrow n = 24 + 1 = 25 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{25} = \frac{25}{2} [2 \times 1 + (25 - 1) 2] = \frac{25}{2} \times 2 [1 + 24] \\ or S_{25} = 25 \times 25 = 625 \end{array}$

Q.38 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?

Ans.

$\begin{array}{l} The given 30 penalties forms an AP with first term as 200 \\ and common difference as 50. \\ So, we have \\ a = 200, d = 50 and n = 30 . \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{30} = \frac{30}{2} [2 \times 200 + (30 - 1) 50] = 15 [400 + 29 \times 50] \\ or S_{30} = 15 [400 + 1450] = 1850 \times 15 = 27, 750 \\ Therefore, the contractor has to pay ₹ 27, 750 as penalty. \end{array}$

Q.39 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.

Ans.

$\begin{array}{l} The given 7 cash prizes forms an AP with common difference \\ as - 20 and sum of all the 7 prizes as 700. \\ So, we have \\ S_{7} = 700, d = - 20 and n = 7 . \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{7} = \frac{7}{2} [2 a + (7 - 1) (- 20)] = \frac{7}{2} \times 2 [a - 60] \\ or 7 00 = 7 [a - 60] \\ or a - 60 = \frac{700}{7} = 100 \\ or a = 100 + 60 = 160 \\ Therefore, the values of all the 7 prizes are ₹ 160, ₹ 140, ₹ 120, \\ ₹ 100, ₹ 80, ₹ 60 and ₹ 40 . \end{array}$

Q.40 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?

Ans.

$\begin{array}{l} It is given that a section of Class I will plant 1 tree, a section \\ of Class II will plant 2 trees and so on till Class XII . \\ The number of trees planted by a section of each class \\ forms an AP with a = 1 and d = 1 . \\ So, we have \\ a = 1, d = 1 and n = 12 . \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{12} = \frac{12}{2} [2 \times 1 + (12 - 1) 1] = 6 [2 + 11] \\ or S_{12} = 78 \\ Therefore, \\ Sum of trees planted by a section of each class = 78 \\ \Rightarrow Sum of trees planted by 3 sections of each class = 78 \times 3 \\ = 234 \end{array}$

Q.41 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 the following figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π =22/7)

[Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, . . . with centres at A, B, A, B, . . ., respectively.]

Ans.

$\begin{array}{l} Here, successive semicircles are of raddi 0.5 cm, 1.0 cm, \\ 1.5 cm, 2.0 cm, . \dots \\ Let l_{1}, l_{2}, l_{3}, l_{4}, . . . are the lengths of successive semicircles \\ in the spiral. \\ We know that length of semicircle is π r. \\ So, lengths of successive semicircles are 0.5 π cm, π cm, \\ 1.5 π cm, 2 π cm, . \dots \\ Sum of lengths of 13 semicircles is π (0.5 + 1 .0 + 1 .5 + 2 .0 + \\ . .. to 13th terms) cm. \\ Now, \\ List of numbers involved in sum 0.5 + 1 .0 + 1 .5 + 2 .0 + . .. \\ to 13th terms forms an AP with a = 0 .5 and d = 0 .5 \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{13} = \frac{13}{2} [2 \times 0 .5 + (13 - 1) (0 .5)] = \frac{13}{2} \times 0 .5 [2 + 12] \\ or S_{13} = \frac{13}{2} \times 0 .5 \times 14 = 45 .5 \\ Therefore, \\ Sum of lengths of 13 semicircles \\ = π (0.5 + 1 .0 + 1 .5 + 2 .0 + . .. to 13th terms) cm \\ = 45 .5 π cm \\ = 45 .5 \times \frac{22}{7} cm \\ = 143 cm \\ Therefore, total length of the spiral is 143 cm . \end{array}$

Q.42 200 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 the following figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Ans.

$\begin{array}{l} It is given that 200 logs are stacked in such a way that \\ 20 logs are in the bottom row, 19 logs are in the next row, \\ 18 \log s are in the row next to it and so on . \\ List of numbers 20, 19, 18, . .. involved in stacking of logs \\ form an AP with a = 20, d = - 1 and S_{n} = 200 . \\ We know that sum of first n terms of an AP is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, 200 = \frac{n}{2} [2 \times 20 + (n - 1) (- 1)] = \frac{n}{2} [40 + - n + 1] \\ or 200 = \frac{n}{2} [41 - n] \\ or 400 = 41 n - n^{2} \\ or n^{2} - 41 n + 400 = 0 \\ or n^{2} - 16 n - 25 n + 400 = 0 \\ or n (n - 16) - 25 (n - 16) = 0 \\ or (n - 16) (n - 25) = 0 \\ or n = 16, n = 25 \\ Now, \\ a_{25} = a + 24 d = 20 + 24 \times (- 1) = - 4 \\ But, number of logs can not be negative. \\ Therefore, number of rows of logs can not be 25 i.e., n \neq 25 . \\ Also, \\ a_{16} = a + 15 d = 20 + 15 \times (- 1) = 5 \\ Therefore, 200 logs are placed in 16 rows and 5 logs are \\ placed in the top row. \end{array}$

Q.43 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 the following figure)

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?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Ans.

$\begin{array}{l} To pick up the first potato and the second potato, the total \\ distance (in metres) run by a competitor is 2×5+2× (5 + 3) . \\ Similarly, the total distance (in metres) run by a competitor \\ to pick up 10 potatoes is 2×5+2× (5+3) +2× (5+3+3) + . .. up to \\ 10th terms. \\ Now, \\ Total distance = 2×5+2× (5+3) +2× (5+3+3) + . .. up to 10th terms. \\ = 2 (5+8+11+ . .. up to 10th terms) \\ = 2× \frac{10}{2} [2×5+(10-1)3] [S_{n} = \frac{n}{2} [2a+(n-1)d]] \\ = 10 [10+27] \\ = 370 m \end{array}$

Q.44 Which term of the AP: 121, 117, 113, . . ., is its first negative term?
[Hint: Find n for a_n < 0]

Ans.

$\begin{array}{l} The given AP is: 121, 117, 113, . . . \\ Here, a = 121, d = 117 - 121 = - 4 \\ Now, \\ a_{n} = a + (n - 1) d \\ = 121 + (n - 1) (- 4) \\ = 121 - 4 n + 4 \\ = 125 - 4 n \\ We have to find the first negative term of this A.P. \\ So, \\ a_{n} < 0 \\ \Rightarrow 125 - 4 n < 0 \\ \Rightarrow 4 n > 125 \\ \Rightarrow n > \frac{125}{4} \\ \Rightarrow n > 31 .25 \\ ∴ n = 32 \\ Hence, 32^{nd} term of the given AP is first negative term. \end{array}$

Q.45 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.

Ans.

$\begin{array}{l} We have \\ a_{3} + a_{7} = 6, a_{3} \times a_{7} = 8 \\ {We know that n}^{th} terms of an AP is given by \\ a_{n} = a + (n - 1) d \\ So, \\ a_{3} = a + 2 d and a_{7} = a + 6 d \\ Now, \\ a_{3} + a_{7} = 6 \\ \Rightarrow a + 2 d + a + 6 d = 6 \\ \Rightarrow 2 a + 8 d = 6 \\ \Rightarrow a + 4 d = 3 \\ \Rightarrow a = 3 - 4 d . .. (1) \\ Again, \\ a_{3} \times a_{7} = 8 \\ \Rightarrow (a + 2 d) \times (a + 6 d) = 8 \\ \Rightarrow (3 - 4 d + 2 d) (3 - 4 d + 6 d) = 8 [From (1)] \\ \Rightarrow (3 - 2 d) (3 + 2 d) = 8 \\ \Rightarrow 9 - 4 d^{2} = 8 \\ \Rightarrow 4 d^{2} = 1 \\ \Rightarrow d = \frac{1}{2} or - \frac{1}{2} \\ On putting this value of d in (1), we get \\ a = 1, when d = \frac{1}{2} \\ and \\ a = 5, when d = - \frac{1}{2} \\ When a = 1 and d = \frac{1}{2} then, \\ S_{16} = \frac{16}{2} [2 \times 1 + 15 \times \frac{1}{2}] = 76 \\ When a = 5 and d = - \frac{1}{2} then, \\ S_{16} = \frac{16}{2} [2 \times 5 + 15 \times (- \frac{1}{2})] = 20 \end{array}$

Q.46

$\begin{array}{l} A ladder has rungs 25 cm apart. (see the following figure) . \\ The rungs decrease uniformly in length from 45 cm at the \\ bottom to 25 cm at the top. If the top and the bottom \\ rungs are 2 \frac{1}{2} m apart, what is the length of the wood \\ required for the rungs? \\ [Hint:Number of rungs= \frac{250}{25}] \end{array}$

Ans.

$\begin{array}{l} It is given that the rungs are 25 cm apart and the top and \\ bottom rungs are \frac{1}{2} m apart. \\ ∴ Total number of rungs= \frac{2 \frac{1}{2} ×100}{25} +1= \frac{250}{25} +1=11 \\ {Lengths of the rungs form an AP with a=45 and a}_{11} =l=25. \\ S_{n} = \frac{n}{2} (a+l) = \frac{11}{2} (45+25) = \frac{11}{2} ×70=385 \\ Therefore, the length of the wood required for the rungs is \\ 385 cm. \end{array}$

Q.47 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.
[Hint: S_x–1 = S₄₉ – S_x]

Ans.

$\begin{array}{l} The houses are numbered consecutively as 1, 2, 3, . .., 49. \\ It can be observed that the number of houses are in an AP \\ having a as 1 and d also as 1. \\ Let there is a number x in this AP such that the sum of the \\ numbers of the houses preceding the house numbered x is \\ equal to the sum of the numbers of the houses following it. \\ We know that, \\ Sum of first n terms of an A.P is given by \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ ∴ {Sum of number of houses preceding x}^{th} house \\ = S_{x - 1} \\ = \frac{x - 1}{2} [2 \times 1 + (x - 1 - 1) 1] \\ = \frac{x - 1}{2} [2 + (x - 2)] \\ = \frac{x (x - 1)}{2} \\ {Sum of number of houses following x}^{th} house \\ = S_{49} - S_{x} \\ = \frac{49}{2} [2 + (49 - 1) 1] - \frac{x}{2} [2 + (x - 1) 1] \\ = 49 \times 25 - \frac{x}{2} (1 + x) \\ It is given that \\ S_{x - 1} = S_{49} - S_{x} \\ \Rightarrow \frac{x (x - 1)}{2} = 49 \times 25 - \frac{x}{2} (1 + x) \\ \Rightarrow x^{2} - x = 2450 - x - x^{2} \\ \Rightarrow 2 x^{2} = 2450 \\ \Rightarrow x^{2} = 1225 \\ \Rightarrow x = \pm 35 \\ But, number of houses can not be negative. \\ ∴ x = 35 \end{array}$

Q.48

$\begin{array}{l} A small terrace at a football ground comprises of 15 \\ steps each of which is 50 m long and built of solid \\ concrete. Each step has a rise of \frac{1}{4} m and a tread of \\ \frac{1}{2} m. (see the following figure) . Calculate the total \\ volume of concrete required to build the terrace. \\ [Hint: Volume of concrete required to build the first \\ step= \frac{1}{4} × \frac{1}{2} {×50 m}^{3}] \end{array}$

Ans.

$\begin{array}{l} According to question, widths of successive steps in meter \\ from top are \frac{1}{2}, 1, \frac{3}{2}, 2, . .. which form an AP with a = d = \frac{1}{2} . \\ Each step is 50 m long and has a rise of \frac{1}{4} m . \\ Now, \\ Volume of concrete required to build the 1st step \\ = \frac{1}{4} \times \frac{1}{2} \times {50 m}^{3} = \frac{25}{4} m^{3} \\ Volume of concrete required to build the 2nd step \\ = \frac{1}{4} \times 1 \times {50 m}^{3} = \frac{25}{2} m^{3} \\ Volume of concrete required to build the 3rd step \\ = \frac{1}{4} \times \frac{3}{2} \times {50 m}^{3} = \frac{75}{4} m^{3} \\ List of numbers \frac{25}{4}, \frac{25}{2}, \frac{75}{4}, . .. form an AP with a = \frac{25}{4} and \\ d = \frac{25}{2} - \frac{25}{4} = \frac{25}{4} . \\ We know that \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ So, S_{15} = \frac{15}{2} [2 \times \frac{25}{4} + (15 - 1) \frac{25}{4}] \\ or S_{15} = \frac{15}{2} [\frac{25}{2} + 14 \times \frac{25}{4}] = \frac{15}{2} [\frac{25}{2} + \frac{175}{2}] = \frac{15}{2} [100] = 750 \\ Therefore, \\ Volume of concrete required to build the terrace is 750 m^{3} . \end{array}$

Q.49

$\begin{array}{l} In which of the following situations, does the list of \\ numbers involved make an arithmetic progression, \\ and why? \\ (i) The taxi fare after each km when the fare is ₹ 15 \\ for the first km and ₹ 8 for each additional km. \\ (ii) The amount of air present in a cylinder when a \\ vacuum pump removes \frac{1}{4} of the air remaining \\ in the cylinder at a time. \\ (iii) The cost of digging a well after every metre of \\ digging, when it costs ₹ 150 for the first metre \\ and rises by ₹ 50 for each subsequent metre. \\ (iv) The amount of money in the account every year, \\ when ₹ 10000 is deposited at compound interest \\ at 8 % per annum. \end{array}$

Ans.

$\begin{array}{l} (i) \\ Fare for the first km = ₹ 15 \\ Fare for the second km = ₹ 15 + ₹ 8 = ₹ 23 \\ Fare for the third km = ₹ 15 + ₹ 8 + ₹ 8 = ₹ 31 \\ and so on. \\ Therefore, fare for each successive kilometre in rupees \\ are: \\ 15, 23, 31, 39, . .. \\ Difference between fares for a particular kilometre and its \\ preceding kilometre is ‘ 8. \\ Therefore, 15, 23, 31, 39, . .. forms an AP. \\ (ii) \\ Let the amount of air present in the cylinder initially is V. \\ The vaccum pump removes \frac{1}{4} of air remaining in the \\ cylinder each time. \\ Therefore, amounts of air in the cylinder after each \\ successive removal by the vacuum pump will be \\ V, \frac{3V}{4}, (\frac{3V}{4} - \frac{3V}{4} \times \frac{1}{4}), \{(\frac{3V}{4} - \frac{3V}{4} \times \frac{1}{4}) - (\frac{3V}{4} - \frac{3V}{4} \times \frac{1}{4}) \times \frac{1}{4}\}, . .. \\ i.e., V, \frac{3V}{4}, {(\frac{3}{4})}^{2} V, {(\frac{3}{4})}^{3} V, . .. \\ Here, the differences between a term and its preceding term \\ are not equal. Therefore, list of numbers involved does \\ not make an A.P. \\ (iii) \\ According to question, cost in rupees of digging first \\ metre and thereafter each subsequent metre are \\ respectively: \\ 150, 150+50, 150+50+50, 150+50+50+50, . .. \\ 150, 200, 250, 300, . .. \\ Differences between a term and its preceding term in the \\ above list of numbers are equal. \\ Therefore, 150, 200, 250, 300, . .. \\ forms an AP. \\ (iv) \\ Amount at the end of first year=10000 {(1+ \frac{8}{100})}^{1} = ₹ 10800 \\ Amount at the end of second year=10800 {(1+ \frac{8}{100})}^{1} = ₹ 11664 \\ Amount at the end of third year=11664 {(1+ \frac{8}{100})}^{1} = ₹ 12597.12 \\ Obviously, differences between amounts in any two successive \\ years are not equal. Therefore, list of numbers involved here \\ does not make an AP. \end{array}$

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NCERT Solutions for Class 10 Maths Related Chapters

Real Numbers

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

Triangles

Coordinate Geometry

Introduction to Trigonometry

Some Applications of Trigonometry

Circles

Constructions

Areas Related to Circles

Surface Areas and Volumes

Statistics

Probability

FAQs (Frequently Asked Questions)

The main nth term formula is aₙ = a + (n – 1)d. The main sum formula is Sₙ = n / 2 [2a + (n – 1)d].

Chapter 5 has Exercise 5.1, 5.2, 5.3 and optional Exercise 5.4. The main exam-focused questions come from AP identification, nth term and sum.

Find the difference between consecutive terms. If the difference remains the same throughout, the sequence is an AP.

Penalty, prizes, trees, spiral, logs and potato race questions are important. They test AP sum formulas in word problems.

Students often confuse nth term with sum of terms. Word problems asking total amount, total distance or total production need the sum formula.

NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

Key Takeaways

NCERT Solutions Class 10 Maths Chapter 5 Structure 2026-27

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise

Class 10 Maths Chapter 5 Exercise 5.1 Solutions

Q1. In which situations does the list of numbers make an arithmetic progression?

(i) Taxi fare after each km

(ii) Air remaining in a cylinder

(iii) Cost of digging a well after every metre

(iv) Amount at compound interest

Q2. Write the first four terms of the AP.

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

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

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

Q3. For the following APs, write the first term and common difference.

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

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

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

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

Q4. Which of the following are APs?

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

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

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

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

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ...

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

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

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

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

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

(xi) a, a², a³, a⁴, ...

(xii) 2, 8, 18, 32, ...

(xiii) √3, √6, √9, √12, ...

(xiv) 1², 3², 5², 7², ...

(xv) 1², 5², 7², 7³, ...

Class 10 Maths Chapter 5 Exercise 5.2 Solutions

Q1. Fill in the blanks in the table.

Q2. Choose the correct answer.

(i) 30th term of 10, 7, 4, ...

(ii) 11th term of -3, -1 / 2, 2, ...

Q3. Find the missing terms.

(i) 2, __, 26

(ii) __, 13, __, 3

(iii) 5, __, __, 9½

(iv) -4, __, __, __, __, 6

(v) __, 38, __, __, __, -22

Q4. Which term of 3, 8, 13, 18, ... is 78?

Q5. Find the number of terms.

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

(ii) 18, 15½, 13, ..., -47

Q6. Check whether -150 is a term of 11, 8, 5, 2, ...

Q7. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Q8. An AP has 50 terms. Its 3rd term is 12 and last term is 106. Find the 29th term.

Q9. If the 3rd and 9th terms are 4 and -8, which term is zero?

Q10. The 17th term exceeds the 10th term by 7. Find d.

Q11. Which term of 3, 15, 27, 39, ... is 132 more than its 54th term?

Q12. Two APs have the same common difference. Their 100th terms differ by 100. Find the difference between their 1000th terms.

Q13. How many three-digit numbers are divisible by 7?

Q14. How many multiples of 4 lie between 10 and 250?

Q15. For what value of n are nth terms of two APs equal?

Q16. Determine the AP whose third term is 16 and 7th term exceeds 5th term by 12.

Q17. Find the 20th term from the last term of 3, 8, 13, ..., 253.

Q18. Sum of 4th and 8th terms is 24. Sum of 6th and 10th terms is 44. Find first three terms.

Q19. Subba Rao started work in 1995 at Rs 5000 salary. Increment is Rs 200 each year. In which year did income reach Rs 7000?

Q20. Ramkali saved Rs 5 in first week and increased weekly savings by Rs 1.75. In nth week, savings became Rs 20.75. Find n.

Class 10 Maths Chapter 5 Exercise 5.3 Solutions

Q1. Find the sum of the following APs.

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

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

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

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

Q2. Find the sums.

(i) 7 + 10½ + 14 + ... + 84

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

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

Q3. Find the missing values.

(i) a = 5, d = 3, aₙ = 50

(ii) a = 7, a₁₃ = 35

(ii) , 13, , 3

(iii) 5, , , 9½

(iv) -4, , , , , 6

(v) , 38, , , , -22