CLASS 10 · MATHEMATICS · CHAPTER 5

Arithmetic Progressions

Recognise arithmetic progressions, calculate any term and use AP sums in number and real-life problems.

98 worked answersAll 4 exercise sets

What you will learn

Test whether a sequence is an AP.
Use the nth-term formula.
Find finite AP sums and solve applications.

Read the NCERT questions ↗← All chapters

98 worked answers · short steps first, explanations when you need them

Exercise 5.1

Textbook pages 55–56

Question 1(i)

A taxi charges ₹15 for the first kilometre and ₹8 for each additional kilometre. Do the fares after successive kilometres form an AP?

Steps only

Fares: 15, 23, 31, 39, …

Consecutive difference = ₹8

Answer

Yes; it is an AP with a = 15 and d = 8.

Question 1(ii)

A pump removes one fourth of the air remaining in a cylinder each time. Do the successive amounts of air form an AP?

Steps only

Amounts: A, 3A/4, 9A/16, 27A/64, …

Differences are −A/4, −3A/16, −9A/64, … and are unequal

Answer

No; the remaining air does not form an AP.

Question 1(iii)

Digging costs ₹150 for the first metre and ₹50 more for every later metre. Do the successive metre costs form an AP?

Steps only

Costs: 150, 200, 250, 300, …

Consecutive difference = ₹50

Answer

Yes; it is an AP with a = 150 and d = 50.

Question 1(iv)

₹10,000 earns compound interest at 8% yearly. Do the yearly account balances form an AP?

Steps only

Balances: 10000, 10800, 11664, …

First increases: 800 and 864, which are unequal

Answer

No; compound-interest balances do not form an AP.

Question 2(i)

Write four terms when a = 10 and d = 10.

Steps only

10, 10 + 10, 10 + 20, 10 + 30

Answer

10, 20, 30, 40.

Question 2(ii)

Write four terms when a = −2 and d = 0.

Steps only

−2, −2 + 0, −2 + 0, −2 + 0

Answer

−2, −2, −2, −2.

Question 2(iii)

Write four terms when a = 4 and d = −3.

Steps only

4, 4 − 3, 4 − 6, 4 − 9

Answer

4, 1, −2, −5.

Question 2(iv)

Write four terms when a = −1 and d = 1/2.

Steps only

−1, −1 + 1/2, −1 + 1, −1 + 3/2

Answer

−1, −1/2, 0, 1/2.

Question 2(v)

Write four terms when a = −1.25 and d = −0.25.

Steps only

−1.25, −1.50, −1.75, −2.00

Answer

−1.25, −1.50, −1.75, −2.00.

Question 3(i)

For 3, 1, −1, −3, … find a and d.

Steps only

a = 3

d = 1 − 3 = −2

Answer

a = 3 and d = −2.

Question 3(ii)

For −5, −1, 3, 7, … find a and d.

Steps only

a = −5

d = −1 − (−5) = 4

Answer

a = −5 and d = 4.

Question 3(iii)

For 1/3, 5/3, 9/3, 13/3, … find a and d.

Steps only

a = 1/3

d = 5/3 − 1/3 = 4/3

Answer

a = 1/3 and d = 4/3.

Question 3(iv)

For 0.6, 1.7, 2.8, 3.9, … find a and d.

Steps only

a = 0.6

d = 1.7 − 0.6 = 1.1

Answer

a = 0.6 and d = 1.1.

Question 4(i)

Decide whether 2, 4, 8, 16, … is an AP; if it is, give d and three more terms.

Steps only

2, 4, 8

Answer

No; differences 2, 4 and 8 are unequal.

Question 4(ii)

Decide whether 2, 5/2, 3, 7/2, … is an AP; if it is, give d and three more terms.

Steps only

1/2

4, 9/2, 5

Answer

Yes; d = 1/2. Next terms: 4, 9/2, 5.

Question 4(iii)

Decide whether −1.2, −3.2, −5.2, −7.2, … is an AP; if it is, give d and three more terms.

Steps only

−2

−9.2, −11.2, −13.2

Answer

Yes; d = −2. Next terms: −9.2, −11.2, −13.2.

Question 4(iv)

Decide whether −10, −6, −2, 2, … is an AP; if it is, give d and three more terms.

Steps only

6, 10, 14

Answer

Yes; d = 4. Next terms: 6, 10, 14.

Question 4(v)

Decide whether 3, 3 + √2, 3 + 2√2, 3 + 3√2, … is an AP; if it is, give d and three more terms.

Steps only

√2

3 + 4√2, 3 + 5√2, 3 + 6√2

Answer

Yes; d = √2. Next terms: 3 + 4√2, 3 + 5√2, 3 + 6√2.

Question 4(vi)

Decide whether 0.2, 0.22, 0.222, 0.2222, … is an AP; if it is, give d and three more terms.

Steps only

0.02, 0.002, 0.0002

Answer

No; the differences are unequal.

Question 4(vii)

Decide whether 0, −4, −8, −12, … is an AP; if it is, give d and three more terms.

Steps only

−4

−16, −20, −24

Answer

Yes; d = −4. Next terms: −16, −20, −24.

Question 4(viii)

Decide whether −1/2, −1/2, −1/2, −1/2, … is an AP; if it is, give d and three more terms.

Steps only

−1/2, −1/2, −1/2

Answer

Yes; d = 0. Next terms are all −1/2.

Question 4(ix)

Decide whether 1, 3, 9, 27, … is an AP; if it is, give d and three more terms.

Steps only

2, 6, 18

Answer

No; the differences are unequal.

Question 4(x)

Decide whether a, 2a, 3a, 4a, … is an AP; if it is, give d and three more terms.

Steps only

5a, 6a, 7a

Answer

Yes; d = a. Next terms: 5a, 6a, 7a.

Question 4(xi)

Decide whether a, a², a³, a⁴, … is an AP; if it is, give d and three more terms.

Steps only

a² − a, a³ − a²

Answer

Not in general; the consecutive differences depend on the term.

Question 4(xii)

Decide whether 2, 8, 18, 32, … is an AP; if it is, give d and three more terms.

Steps only

6, 10, 14

Answer

No; the differences are unequal.

Question 4(xiii)

Decide whether √3, √6, √9, √12, … is an AP; if it is, give d and three more terms.

Steps only

√6 − √3, 3 − √6

Answer

No; the differences are unequal.

Question 4(xiv)

Decide whether 1², 3², 5², 7², … is an AP; if it is, give d and three more terms.

Steps only

8, 16, 24

Answer

No; the differences are unequal.

Question 4(xv)

Decide whether 1², 5², 7², 73, … is an AP; if it is, give d and three more terms.

Steps only

24, 24, 24

Answer

Yes; its values are 1, 25, 49, 73, so d = 24. Next terms: 97, 121, 145.

Exercise 5.2

Textbook pages 61–63

Question 1(i)

Given a = 7, d = 3 and n = 8, find a₈.

Steps only

a₈ = 7 + 7(3) = 28

Answer

a₈ = 28.

Question 1(ii)

Given a = −18, n = 10 and a₁₀ = 0, find d.

Steps only

0 = −18 + 9d

d = 2

Answer

d = 2.

Question 1(iii)

Given d = −3, n = 18 and a₁₈ = −5, find a.

Steps only

−5 = a + 17(−3)

a = 46

Answer

a = 46.

Question 1(iv)

Given a = −18.9, d = 2.5 and aₙ = 3.6, find n.

Steps only

3.6 = −18.9 + (n − 1)2.5

n − 1 = 9, so n = 10

Answer

n = 10.

Question 1(v)

Given a = 3.5, d = 0 and n = 105, find a₁₀₅.

Steps only

a₁₀₅ = 3.5 + 104(0)

Answer

a₁₀₅ = 3.5.

Question 2(i)

Find the 30th term of 10, 7, 4, ….

Steps only

a = 10, d = −3

a₃₀ = 10 + 29(−3) = −77

Answer

Option C: −77.

Question 2(ii)

Find the 11th term of −3, −1/2, 2, ….

Steps only

d = −1/2 − (−3) = 5/2

a₁₁ = −3 + 10(5/2) = 22

Answer

Option B: 22.

Question 3(i)

Complete 2, □, 26.

Steps only

d = (26 − 2)/2 = 12

Middle term = 2 + 12 = 14

Answer

The missing term is 14.

Question 3(ii)

Complete □, 13, □, 3.

Steps only

a + d = 13; a + 3d = 3

2d = −10 ⇒ d = −5; a = 18

a₃ = 18 − 10 = 8

Answer

The missing terms are 18 and 8.

Question 3(iii)

Complete 5, □, □, 9 1/2.

Steps only

5 + 3d = 19/2 ⇒ d = 3/2

a₂ = 13/2; a₃ = 8

Answer

The missing terms are 13/2 and 8.

Question 3(iv)

Complete −4, □, □, □, □, 6.

Steps only

d = [6 − (−4)]/5 = 2

Terms: −4, −2, 0, 2, 4, 6

Answer

The missing terms are −2, 0, 2 and 4.

Question 3(v)

Complete □, 38, □, □, □, −22.

Steps only

38 + 4d = −22 ⇒ d = −15

a = 38 − d = 53

Terms: 53, 38, 23, 8, −7, −22

Answer

The missing terms are 53, 23, 8 and −7.

Question 4

Which term of 3, 8, 13, 18, … is 78?

Steps only

78 = 3 + 5(n − 1)

n = 16

Answer

78 is the 16th term.

Question 5(i)

Find the number of terms in 7, 13, 19, …, 205.

Steps only

205 = 7 + 6(n − 1)

n = 34

Answer

There are 34 terms.

Question 5(ii)

Find the number of terms in 18, 15 1/2, 13, …, −47.

Steps only

a = 18, d = −5/2

−47 = 18 − (5/2)(n − 1)

n = 27

Answer

There are 27 terms.

Question 6

Is −150 a term of 11, 8, 5, 2, …?

Steps only

−150 = 11 − 3(n − 1)

n = 164/3, not a positive integer

Answer

No, −150 is not a term.

Question 7

An AP has a₁₁ = 38 and a₁₆ = 73. Find a₃₁.

Steps only

5d = 73 − 38 = 35 ⇒ d = 7

a = 38 − 10(7) = −32

a₃₁ = −32 + 30(7) = 178

Answer

The 31st term is 178.

Question 8

A 50-term AP has a₃ = 12 and a₅₀ = 106. Find a₂₉.

Steps only

a + 2d = 12; a + 49d = 106

47d = 94 ⇒ d = 2 and a = 8

a₂₉ = 8 + 28(2) = 64

Answer

The 29th term is 64.

Question 9

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

Steps only

a₁₇ − a₁₀ = 7d

7d = 7

Answer

d = 1.

Question 10

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

Steps only

aₙ − a₅₄ = (n − 54)12 = 132

n − 54 = 11

Answer

The 65th term.

Question 11

Two APs have the same common difference and their 100th terms differ by 100. Find the difference of their 1000th terms.

Steps only

Difference at term n = (a − b) + (n − 1)(d − d)

= a − b, so it is constant

Answer

The difference is 100.

Question 12

How many three-digit numbers are divisible by 7?

Steps only

First = 105; last = 994; d = 7

994 = 105 + 7(n − 1)

n = 128

Answer

There are 128.

Question 13

How many multiples of 4 lie strictly between 10 and 250?

Steps only

Sequence: 12, 16, …, 248

248 = 12 + 4(n − 1)

n = 60

Answer

There are 60 multiples.

Question 14

For what n are the nth terms of 63, 65, 67, … and 3, 10, 17, … equal?

Steps only

First AP: aₙ = 63 + 2(n − 1) = 61 + 2n

Second AP: aₙ = 3 + 7(n − 1) = 7n − 4

61 + 2n = 7n − 4 ⇒ n = 13

Answer

n = 13.

Question 15

The third term is 16 and the seventh exceeds the fifth by 12. Determine the AP.

Steps only

a + 2d = 16

a₇ − a₅ = 2d = 12 ⇒ d = 6

a = 4

Answer

The AP is 4, 10, 16, 22, ….

Question 16

The 3rd and 9th terms are 4 and −8. Which term is zero?

Steps only

a + 2d = 4; a + 8d = −8

d = −2 and a = 8

0 = 8 − 2(n − 1) ⇒ n = 5

Answer

The 5th term is zero.

Question 17

Find the 20th term from the end of 3, 8, 13, …, 253.

Steps only

Reverse AP: a = 253, d = −5

a₂₀ = 253 + 19(−5) = 158

Answer

The required term is 158.

Question 18

a₄ + a₈ = 24 and a₆ + a₁₀ = 44. Find the first three terms.

Steps only

2a + 10d = 24; 2a + 14d = 44

4d = 20 ⇒ d = 5; a = −13

Answer

The first three terms are −13, −8 and −3.

Question 19

A salary was ₹5000 in 1995 and rose ₹200 yearly. In which year did it reach ₹7000?

Steps only

7000 = 5000 + 200(n − 1)

n = 11

1995 + 10 = 2005

Answer

It reached ₹7000 in 2005.

Question 20

Savings start at ₹5 and rise ₹1.75 weekly. If the nth-week saving is ₹20.75, find n.

Steps only

20.75 = 5 + 1.75(n − 1)

n − 1 = 9

Answer

n = 10.

Exercise 5.3

Textbook pages 68–70

Question 1(i)

Find the sum of 2, 7, 12, … to 10 terms.

Steps only

S₁₀ = 10[2(2) + 9(5)]/2

= 5(49) = 245

Answer

245.

Question 1(ii)

Find the sum of −37, −33, −29, … to 12 terms.

Steps only

S₁₂ = 12[2(−37) + 11(4)]/2

= 6(−30) = −180

Answer

−180.

Question 1(iii)

Find the sum of 0.6, 1.7, 2.8, … to 100 terms.

Steps only

a = 0.6, d = 1.1

S₁₀₀ = 50[1.2 + 99(1.1)] = 5505

Answer

5505.

Question 1(iv)

Find the sum of 1/15, 1/12, 1/10, … to 11 terms.

Steps only

d = 1/60

S₁₁ = 11[2/15 + 10/60]/2

= 33/20

Answer

33/20.

Question 2(i)

Find 7 + 10 1/2 + 14 + … + 84.

Steps only

a = 7, d = 7/2, l = 84

84 = 7 + (n − 1)(7/2) ⇒ n = 23

S = 23(7 + 84)/2 = 2093/2

Answer

2093/2 or 1046.5.

Question 2(ii)

Find 34 + 32 + 30 + … + 10.

Steps only

a = 34, d = −2, l = 10, n = 13

S = 13(34 + 10)/2 = 286

Answer

286.

Question 2(iii)

Find −5 + (−8) + (−11) + … + (−230).

Steps only

a = −5, d = −3, l = −230, n = 76

S = 76(−5 − 230)/2 = −8930

Answer

−8930.

Question 3(i)

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

Steps only

50 = 5 + 3(n − 1) ⇒ n = 16

S₁₆ = 16(5 + 50)/2 = 440

Answer

n = 16 and Sₙ = 440.

Question 3(ii)

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

Steps only

35 = 7 + 12d ⇒ d = 7/3

S₁₃ = 13(7 + 35)/2 = 273

Answer

d = 7/3 and S₁₃ = 273.

Question 3(iii)

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

Steps only

37 = a + 11(3) ⇒ a = 4

S₁₂ = 12(4 + 37)/2 = 246

Answer

a = 4 and S₁₂ = 246.

Question 3(iv)

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

Steps only

a + 2d = 15

5(2a + 9d) = 125 ⇒ d = −1, a = 17

a₁₀ = 17 − 9 = 8

Answer

d = −1 and a₁₀ = 8.

Question 3(v)

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

Steps only

75 = 9[2a + 8(5)]/2

a = −35/3

a₉ = a + 40 = 85/3

Answer

a = −35/3 and a₉ = 85/3.

Question 3(vi)

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

Steps only

90 = n[4 + 8(n − 1)]/2

2n² − n − 45 = 0 ⇒ n = 5

a₅ = 2 + 4(8) = 34

Answer

n = 5 and aₙ = 34.

Question 3(vii)

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

Steps only

210 = n(8 + 62)/2 = 35n ⇒ n = 6

62 = 8 + 5d ⇒ d = 54/5

Answer

n = 6 and d = 54/5.

Question 3(viii)

Given aₙ = 4, d = 2 and Sₙ = −14, find n and a.

Steps only

a = 4 − 2(n − 1) = 6 − 2n

−14 = n(a + 4)/2 = n(5 − n)

n² − 5n − 14 = 0 ⇒ n = 7; a = −8

Answer

n = 7 and a = −8.

Question 3(ix)

Given a = 3, n = 8 and S₈ = 192, find d.

Steps only

192 = 8[6 + 7d]/2

48 = 6 + 7d ⇒ d = 6

Answer

d = 6.

Question 3(x)

A nine-term AP has last term 28 and sum 144. Find a.

Steps only

144 = 9(a + 28)/2

a + 28 = 32

Answer

a = 4.

Question 4

How many terms of 9, 17, 25, … have sum 636?

Steps only

636 = n[18 + 8(n − 1)]/2

4n² + 5n − 636 = 0

(n − 12)(4n + 53) = 0

Answer

12 terms.

Question 5

An AP has first term 5, last term 45 and sum 400. Find n and d.

Steps only

400 = n(5 + 45)/2 ⇒ n = 16

45 = 5 + 15d ⇒ d = 8/3

Answer

n = 16 and d = 8/3.

Question 6

An AP has first term 17, last term 350 and d = 9. Find its number of terms and sum.

Steps only

350 = 17 + 9(n − 1) ⇒ n = 38

S = 38(17 + 350)/2 = 6973

Answer

38 terms; sum = 6973.

Question 7

Find S₂₂ when d = 7 and a₂₂ = 149.

Steps only

a = 149 − 21(7) = 2

S₂₂ = 22(2 + 149)/2 = 1661

Answer

1661.

Question 8

Find S₅₁ if the second and third terms are 14 and 18.

Steps only

d = 18 − 14 = 4; a = 10

S₅₁ = 51[20 + 50(4)]/2 = 5610

Answer

5610.

Question 9

If S₇ = 49 and S₁₇ = 289, find Sₙ.

Steps only

a + 3d = 7; a + 8d = 17

d = 2 and a = 1

Sₙ = n[2 + 2(n − 1)]/2 = n²

Answer

Sₙ = n².

Question 10(i)

Show aₙ = 3 + 4n forms an AP and find S₁₅.

Steps only

a₁ = 7 and aₙ₊₁ − aₙ = 4

S₁₅ = 15[2(7) + 14(4)]/2 = 525

Answer

It is an AP with d = 4; S₁₅ = 525.

Question 10(ii)

Show aₙ = 9 − 5n forms an AP and find S₁₅.

Steps only

a₁ = 4 and aₙ₊₁ − aₙ = −5

S₁₅ = 15[2(4) + 14(−5)]/2 = −465

Answer

It is an AP with d = −5; S₁₅ = −465.

Question 11

If Sₙ = 4n − n², find S₁, S₂, a₂, a₃, a₁₀ and aₙ.

Steps only

S₁ = 3; S₂ = 4

a₂ = S₂ − S₁ = 1; a₃ = S₃ − S₂ = −1

a₁₀ = S₁₀ − S₉ = −15; aₙ = Sₙ − Sₙ₋₁ = 5 − 2n

Answer

S₁ = 3, S₂ = 4, a₂ = 1, a₃ = −1, a₁₀ = −15 and aₙ = 5 − 2n.

Question 12

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

Steps only

Sequence: 6, 12, …, 240

S₄₀ = 40(6 + 240)/2 = 4920

Answer

4920.

Question 13

Find the sum of the first 15 multiples of 8.

Steps only

Sequence: 8, 16, …, 120

S₁₅ = 15(8 + 120)/2 = 960

Answer

960.

Question 14

Find the sum of odd numbers between 0 and 50.

Steps only

Sequence: 1, 3, …, 49; n = 25

S = 25(1 + 49)/2 = 625

Answer

625.

Question 15

A 30-day delay costs ₹200 on day one and ₹50 more each later day. Find the total penalty.

Steps only

a = 200, d = 50, n = 30

S₃₀ = 30[400 + 29(50)]/2 = 27750

Answer

₹27,750.

Question 16

Seven prizes total ₹700 and each is ₹20 less than the preceding prize. Find all prizes.

Steps only

700 = 7[2a + 6(−20)]/2

2a − 120 = 200 ⇒ a = 160

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

Answer

₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.

Question 17

Three sections of each Class I through XII plant as many trees per section as the class number. Find the total.

Steps only

One set of sections: 1 + 2 + … + 12 = 12(13)/2 = 78

Three sections: 3 × 78 = 234

Answer

234 trees.

Question 18

Thirteen semicircles have radii 0.5, 1.0, 1.5, … cm. Find their total arc length using π = 22/7.

Steps only

Sum of radii = 13(0.5 + 6.5)/2 = 45.5 cm

Total semicircle length = πΣr = (22/7)(45.5)

Answer

143 cm.

Question 19

There are 200 logs in rows of 20, 19, 18, … . Find the number of rows and logs in the top row.

Steps only

Top row l = 21 − n

200 = n(20 + 21 − n)/2

n² − 41n + 400 = 0 ⇒ n = 16; l = 5

Answer

16 rows, with 5 logs in the top row.

Question 20

Ten potatoes start 5 m from a bucket and are 3 m apart. Find the total running distance when each is collected separately.

Steps only

One-way distances: 5, 8, 11, …, 32

Sum = 10(5 + 32)/2 = 185 m

Every trip is out and back: 2 × 185 = 370 m

Answer

370 m.

Exercise 5.4 (Optional)

Textbook pages 71

Question 1

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

Steps only

aₙ = 121 − 4(n − 1) = 125 − 4n

125 − 4n < 0 ⇒ n > 31.25

First integer is n = 32; a₃₂ = −3

Answer

The 32nd term is the first negative term.

Question 2

The third and seventh terms have sum 6 and product 8. Find S₁₆.

Steps only

Let their midpoint a₅ = 3: a₃ = 3 − 2d, a₇ = 3 + 2d

9 − 4d² = 8 ⇒ d = ±1/2

If d = 1/2, S₁₆ = 76; if d = −1/2, S₁₆ = 20

Answer

The possible sums are 76 and 20.

Question 3

Rungs 25 cm apart decrease from 45 cm to 25 cm over a vertical distance of 2.5 m. Find the total rung length.

Steps only

Intervals = 250/25 = 10, so rungs = 11

S₁₁ = 11(45 + 25)/2 = 385 cm

Answer

385 cm or 3.85 m.

Question 4

Houses are numbered 1 to 49. Find x so the sum before x equals the sum after x.

Steps only

x(x − 1)/2 = 49(50)/2 − x(x + 1)/2

2x² = 2450

x² = 1225 ⇒ x = 35

Answer

x = 35.

Question 5

A terrace has 15 concrete steps, each 50 m long, rise 1/4 m and tread 1/2 m. Find the total concrete volume.

Steps only

Successive cross-section heights are 1/4, 2/4, …, 15/4 m

Volume = 50(1/2)(1/4)(1 + 2 + … + 15)

= 25/4 × 120 = 750 m³

Answer

750 m³.