In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.

In which of the following situations, does the list of numbers involved make an arithmetic

progression, and why’

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

additional km.

(ii) The amount of air present In a cylinder when a vacuum pump removes 1/4 of the

air remaining in the cylinder at a time.

(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.

Solution:

(i) Let us consider, the first term

(T_{1}) Fare for the first 1 km = ₹ 15

Since, the taxi fare beyond the first 1 km is ₹ 8 for each additional km.

∴ Fare for 2 km = ₹ 15 + 1 × ₹ 8

T_{2} = a + 8 [where a = 15]

Fare for 3 km = ₹ 15 + 2 × ₹ 8

⇒ T_{3} = a + 16

Fare for 4 km= ₹ 15 + 3 × ₹ 8

⇒ T_{4} = a + 24

Fare for 5 km = ₹ 15 + 4 × ₹ 8

⇒ T_{5} = a + 32

Fare for n km = ₹ 15 + (n – 1)8

⇒ T_{n} = a + (n – 1)8

We see that above terms form an AP with common difference 8.

(ii) Let the amount of air present in the cylinder be x

The above terms are not in A.P.

(iii) Here, the cost of digging for first 1 metre = ₹ 150

The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200

The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250

The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300

∴ The terms are: 150, 200, 250, 300,…..

Since, 200 – 150 = 50 and 250 – 200 = 50

(200 – 150) (250 – 200) = 50

∴ The above terms form an AP with common difference 50.

(iv)

Question 2.

Write first four terms of the AP, when the first term a and the common difference d are given

as follows:

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

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

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

Solution:

(i) ∵ T_{n} = a + (n – 1)d

∴ For a = 10 and d = 10, we have:

T_{1} =10 + (1 – 1) × 10 = 10 + 0 = 10

T_{2} = 10 + (2 – 1) × 10 = 10 + 10 – 20

T_{3} = 10 + ( 3 – 1) × 10 = 10 + 20 = 30

T_{4} = 10 + (4 – 1) × 10 = 10 + 30 = 40

Thus, the first four terms are:

10, 20, 30, 40

(ii) ∵ T_{n} = a + (n – 1)d

∴ For a = -2 and d = 0,we have:

T_{1} = -2 + (1 – 1) × 0 = -2 + 0 = -2

T_{2} = -2 + (2 – 1) × 0 = -2 + 0 = -2

T_{3} = -2 + (3 – 1) × 0= -2 + 0 = -2

T_{4} = -2 + (4 – 1) × 0 = -2 + 0 = -2

∴ Thus, the first four terms are: -2, -2, -2, -2

(iii) ∵ T_{n} = a + (n – 1)d

For a = 4 and d = -3, we have:

T_{1} = 4 + (1 – 1) × (-3) = 4 + 0 = 4

T_{2} = 4 + (2 – 1) × (-3) = 4 + (-3) = 1

T_{3} = 4 + (3 – 1) × (-3) = 4 + (-6) = -2

T_{4} = 4 + (4 – 1) × (-3) = 4 + (-9) = -5

Thus, the first four terms are:

4, 1, -2, -5

(iv)

(v) ∵ T_{n} = a + (n – 1)d

∴ For a = -1.25 and d = -0.25, we have

T_{1} = -1.25 + (1 – 1) × (-0.25) = -1.25 + 0

= -1.25

T_{2} = -1.25 + (2 – 1) × (-0.25) = -1.25 + (-0.25) = -1.50

T_{3} = -1.25 + (3 – 1) × (-0.25) = -1.25 + (-0.50) = -1.75

T_{4} = -1.25 + (4 – 1) × (-0.25) = -1.25 + (-0.75) = -2.0

Question 3.

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

(i) 3, 1, -1, -3,…..

(ii) -5, -1, 3, 7,….

(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots .\)

(iv) 0.6, 1.7, 2.8, 3.9, …..

Solution:

(i) We have : 3, 1, -1, -3, …

⇒ T_{1} ⇒ 3 = a = 3

T_{2} = 1, T_{3}= -1, T_{4}= -3

∴ T_{2} – T_{1} = 1 – 3 = -2

T_{4} = T_{3} = -3 – (-1) = -3 + 1 = -2 ⇒ d = -2

(ii) We have : -5, -1, 3, 7,….

⇒ T_{1} = -5 ⇒ a = -5, T_{2} = -1, T_{3} = 3, T_{4} = 7

∴ T_{2} – T_{1} = -1 – (-5) = -1 + 5 = 4

and T_{4} – T_{3} = 7 – 3 = 4 ⇒ d = 4

Thus a = -5, d = 4

(iii)

Question 4.

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

(i) 2, 4, 8, 16,

(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ….

(iii) -1.2, -3.2, -5.2, -7.2, ……

(iv) -10, -6, -2, 2, ….

(v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots\)

(vi) 0.2, 0.22, 0.222, 0.2222, ….

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

(ix) 1, 3, 9, 27, …..

(x) o, 2a, 3a, 4a,…

(xi) a, a_{2}, a_{3}, a_{4},….

(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)

(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots .\)

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2}, ….

(xv) 1^{2}, 5^{2}, 7^{2}, 73, …..

Solution:

(i)

∴ The given numbers do not form an AP

(ii)

(iii) We have : -1.2, -3.2, -5.2, -7.2,

∴ T_{1} = -1.2, T_{2} = -3.2, T_{3} = -5.2, T_{4} = -7.2

T_{2} – T_{1} = -3.2 + 1.2 = -2

T_{3} – T_{2} = -5.2 + 3.2 = -2

T_{4} – T_{3} = -7.2 + 5.2 = -2

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = -2 d = -2

∴ The given numbers form an AP such that d = -2.

Now, T_{5} = T_{4} + (-2) = -7.2 + (-2) = -9.2,

T_{6} = T_{5} + (-2) = -9.2 + (-2) = -11.2 and

T_{7} = T_{6} + (-2) = -11.2 + (-2) = -13.2

Thus, d = -2 and T_{5} = -9.2, T_{5} = -11.2 and T_{6} = -13.2

(iv) We have : -10, -6, -2, 2,

T_{1} = -10, T_{2} = -6, T_{3} = -2, T_{4} = 2

T_{2} – T_{1} = -6 + 10 = 4

T_{3} – T_{2} = -2 + 6 = 4

T_{4} – T_{3} = 2 + 2 = 4

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = 4 ⇒ d = 4

∴ The given numbers form an AP Now, T_{5} = T_{4} + 4 = 2 + 4 = 6,

T_{6} = T_{5} + 4 = 6 + 4 = 10,

T_{7} = T_{6} + 4 = 10 + 4 = 14

Thus, d = 4 and T_{5} = 6, T_{6} = 10, T_{7} = 14

(v) We have :

3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….

(vi) We have : 0.2, 0.22, 0.222, 0.2222,….

∴ The given numbers do not form an AP

(vii) We have : 0, – 4, – 8, – 12,

∴ T_{1} = 0, T_{2} = -4, T_{3} = -8, T_{4} = -12 T_{2} – T_{1} = -4 – 0 = -4

T_{3} – T_{2} = – 8 + 4 = – 4

T_{4} – T_{3} = -12 + 8 = -4

∴ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = -4 ⇒ d = -4

∴ The given numbers form an AP

Now, T_{5} = T_{4} + (- 4) = -12 + (- 4) = -16

T_{6} = T_{5} + (-4) = -16 + (-4)= -20

T_{7} = T_{6} + (- 4) = -20 + (- 4) = -24

Thus, d = – 4 and T_{5} = -16, T_{6} = -20,

T_{7} = -24.

(viii)

(ix) We have, 1, 3, 9, 27,….

(x) We have : a, 2a, 3a, 4a,

(xi)

(xii)

(xiii)

(xiv)

We have : 1^{2}, 3_{2}, 5_{2}, 7_{2},….

(xv) We have : 1_{2} , 5_{2}, 7_{2}, 7_{2},….

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.

In which of the following situations, does the list of numbers involved make an arithmetic

progression, and why’

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

additional km.

(ii) The amount of air present In a cylinder when a vacuum pump removes 1/4 of the

air remaining in the cylinder at a time.

(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.

Solution:

(i) Let us consider, the first term

(T_{1}) Fare for the first 1 km = ₹ 15

Since, the taxi fare beyond the first 1 km is ₹ 8 for each additional km.

∴ Fare for 2 km = ₹ 15 + 1 × ₹ 8

T_{2} = a + 8 [where a = 15]

Fare for 3 km = ₹ 15 + 2 × ₹ 8

⇒ T_{3} = a + 16

Fare for 4 km= ₹ 15 + 3 × ₹ 8

⇒ T_{4} = a + 24

Fare for 5 km = ₹ 15 + 4 × ₹ 8

⇒ T_{5} = a + 32

Fare for n km = ₹ 15 + (n – 1)8

⇒ T_{n} = a + (n – 1)8

We see that above terms form an AP with common difference 8.

(ii) Let the amount of air present in the cylinder be x

The above terms are not in A.P.

(iii) Here, the cost of digging for first 1 metre = ₹ 150

The cost of digging for first 2 metres = ₹ 150 + ₹ 50 = ₹ 200

The cost of digging for first 3 metres = ₹ 150 + (₹ 50) × 2 = ₹ 250

The cost of digging for first 4 metres = ₹ 150 + (₹ 50) × 3 = ₹ 300

∴ The terms are: 150, 200, 250, 300,…..

Since, 200 – 150 = 50 and 250 – 200 = 50

(200 – 150) (250 – 200) = 50

∴ The above terms form an AP with common difference 50.

(iv)

Question 2.

Write first four terms of the AP, when the first term a and the common difference d are given

as follows:

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

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

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

Solution:

(i) ∵ T_{n} = a + (n – 1)d

∴ For a = 10 and d = 10, we have:

T_{1} =10 + (1 – 1) × 10 = 10 + 0 = 10

T_{2} = 10 + (2 – 1) × 10 = 10 + 10 – 20

T_{3} = 10 + ( 3 – 1) × 10 = 10 + 20 = 30

T_{4} = 10 + (4 – 1) × 10 = 10 + 30 = 40

Thus, the first four terms are:

10, 20, 30, 40

(ii) ∵ T_{n} = a + (n – 1)d

∴ For a = -2 and d = 0,we have:

T_{1} = -2 + (1 – 1) × 0 = -2 + 0 = -2

T_{2} = -2 + (2 – 1) × 0 = -2 + 0 = -2

T_{3} = -2 + (3 – 1) × 0= -2 + 0 = -2

T_{4} = -2 + (4 – 1) × 0 = -2 + 0 = -2

∴ Thus, the first four terms are: -2, -2, -2, -2

(iii) ∵ T_{n} = a + (n – 1)d

For a = 4 and d = -3, we have:

T_{1} = 4 + (1 – 1) × (-3) = 4 + 0 = 4

T_{2} = 4 + (2 – 1) × (-3) = 4 + (-3) = 1

T_{3} = 4 + (3 – 1) × (-3) = 4 + (-6) = -2

T_{4} = 4 + (4 – 1) × (-3) = 4 + (-9) = -5

Thus, the first four terms are:

4, 1, -2, -5

(iv)

(v) ∵ T_{n} = a + (n – 1)d

∴ For a = -1.25 and d = -0.25, we have

T_{1} = -1.25 + (1 – 1) × (-0.25) = -1.25 + 0

= -1.25

T_{2} = -1.25 + (2 – 1) × (-0.25) = -1.25 + (-0.25) = -1.50

T_{3} = -1.25 + (3 – 1) × (-0.25) = -1.25 + (-0.50) = -1.75

T_{4} = -1.25 + (4 – 1) × (-0.25) = -1.25 + (-0.75) = -2.0

Question 3.

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

(i) 3, 1, -1, -3,…..

(ii) -5, -1, 3, 7,….

(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots .\)

(iv) 0.6, 1.7, 2.8, 3.9, …..

Solution:

(i) We have : 3, 1, -1, -3, …

⇒ T_{1} ⇒ 3 = a = 3

T_{2} = 1, T_{3}= -1, T_{4}= -3

∴ T_{2} – T_{1} = 1 – 3 = -2

T_{4} = T_{3} = -3 – (-1) = -3 + 1 = -2 ⇒ d = -2

(ii) We have : -5, -1, 3, 7,….

⇒ T_{1} = -5 ⇒ a = -5, T_{2} = -1, T_{3} = 3, T_{4} = 7

∴ T_{2} – T_{1} = -1 – (-5) = -1 + 5 = 4

and T_{4} – T_{3} = 7 – 3 = 4 ⇒ d = 4

Thus a = -5, d = 4

(iii)

Question 4.

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

(i) 2, 4, 8, 16,

(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ….

(iii) -1.2, -3.2, -5.2, -7.2, ……

(iv) -10, -6, -2, 2, ….

(v) \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots\)

(vi) 0.2, 0.22, 0.222, 0.2222, ….

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

(ix) 1, 3, 9, 27, …..

(x) o, 2a, 3a, 4a,…

(xi) a, a_{2}, a_{3}, a_{4},….

(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\)

(xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots .\)

(xiv) 1^{2}, 3^{2}, 5^{2}, 7^{2}, ….

(xv) 1^{2}, 5^{2}, 7^{2}, 73, …..

Solution:

(i)

∴ The given numbers do not form an AP

(ii)

(iii) We have : -1.2, -3.2, -5.2, -7.2,

∴ T_{1} = -1.2, T_{2} = -3.2, T_{3} = -5.2, T_{4} = -7.2

T_{2} – T_{1} = -3.2 + 1.2 = -2

T_{3} – T_{2} = -5.2 + 3.2 = -2

T_{4} – T_{3} = -7.2 + 5.2 = -2

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = -2 d = -2

∴ The given numbers form an AP such that d = -2.

Now, T_{5} = T_{4} + (-2) = -7.2 + (-2) = -9.2,

T_{6} = T_{5} + (-2) = -9.2 + (-2) = -11.2 and

T_{7} = T_{6} + (-2) = -11.2 + (-2) = -13.2

Thus, d = -2 and T_{5} = -9.2, T_{5} = -11.2 and T_{6} = -13.2

(iv) We have : -10, -6, -2, 2,

T_{1} = -10, T_{2} = -6, T_{3} = -2, T_{4} = 2

T_{2} – T_{1} = -6 + 10 = 4

T_{3} – T_{2} = -2 + 6 = 4

T_{4} – T_{3} = 2 + 2 = 4

∵ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = 4 ⇒ d = 4

∴ The given numbers form an AP Now, T_{5} = T_{4} + 4 = 2 + 4 = 6,

T_{6} = T_{5} + 4 = 6 + 4 = 10,

T_{7} = T_{6} + 4 = 10 + 4 = 14

Thus, d = 4 and T_{5} = 6, T_{6} = 10, T_{7} = 14

(v) We have :

3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….

(vi) We have : 0.2, 0.22, 0.222, 0.2222,….

∴ The given numbers do not form an AP

(vii) We have : 0, – 4, – 8, – 12,

∴ T_{1} = 0, T_{2} = -4, T_{3} = -8, T_{4} = -12 T_{2} – T_{1} = -4 – 0 = -4

T_{3} – T_{2} = – 8 + 4 = – 4

T_{4} – T_{3} = -12 + 8 = -4

∴ T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = -4 ⇒ d = -4

∴ The given numbers form an AP

Now, T_{5} = T_{4} + (- 4) = -12 + (- 4) = -16

T_{6} = T_{5} + (-4) = -16 + (-4)= -20

T_{7} = T_{6} + (- 4) = -20 + (- 4) = -24

Thus, d = – 4 and T_{5} = -16, T_{6} = -20,

T_{7} = -24.

(viii)

(ix) We have, 1, 3, 9, 27,….

(x) We have : a, 2a, 3a, 4a,

(xi)

(xii)

(xiii)

(xiv)

We have : 1^{2}, 3_{2}, 5_{2}, 7_{2},….

(xv) We have : 1_{2} , 5_{2}, 7_{2}, 7_{2},….