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These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 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 \(\frac { 1 }{ 4 }\) 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.

Solution:

(i) Yes, 15, 23, 31, …. forms an AP as each succeeding term is obtained by adding 8 in its preceeding term.

(ii) No, volumes are V, \(\frac { 3V }{ 4 }\), (\(\frac { 3V }{ 4 }\))², L.

(iii) Yes, 150,200,250,…. form an A.P.

(iv) No, Amounts are

10000(1 + \(\frac { 8 }{ 100 }\)), 10000(1 + \(\frac { 8 }{ 100 }\))², 10000(1 + \(\frac { 8 }{ 100 }\))², L

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

(n) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d = \(\frac { 1 }{ 2 }\)

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

Solution:

(i) Given: a = 10, d = 10

a_{1} = 10,

a_{2} = 10 + 10 = 20

a_{3} = 20 + 10 = 30

a_{4} = 30 + 10 = 40

Thus, the first four terms of the AP are 10, 20, 30, 40.

(ii) Given: a = – 2, d = 0

The first four terms of the AP are -2, -2, -2, -2.

(iii) a_{1} = 4, d = -3

a_{2} = a_{1} + d = 4 – 3 = 1

a_{3} = a_{2} + d = 1 – 3 = -2

a_{4} = a_{3} + d = -2 – 3 = -5

Thus, the first four terms of the AP are 4, 1, -2, … -5.

(iv)

(v) a_{1} = -1.25, d = -0.25

a_{2} = a_{1} + d = -1.25 – 0.25 = -1.50

a_{3} = a_{2} + d = -1.50 – 0.25 = -1.75

a_{4} = a_{3} + d = -1.75 – 0.25 = -2.00

Thus, the first four terms of the AP are -1.25, -1.50, -1.75, -2.

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

(iv) 0.6, 1.7, 2.8, 3.9, …….

Solution:

(i) a_{1} = 3, a_{2} = 1

d = a_{2} – a_{1} = 1 – 3 = -2

where, a_{1} = first term and d = common difference

a_{1} = 3, d = -2

(ii) a_{1} = -5, a_{2} = -1

d = a_{2} – a_{1} = -1 – (-5) = -1 + 5 = 4

So, first term a_{1} = -5 and common difference d = 4

(iii) a_{1} = \(\frac { 1 }{ 3 }\), a_{2} = \(\frac { 5 }{ 3 }\)

d = \(\frac { 5 }{ 3 }\) – \(\frac { 1 }{ 3 }\) = \(\frac { 4 }{ 3 }\)

So, first term a_{1} = \(\frac { 1 }{ 3 }\) and common difference d = \(\frac { 4 }{ 3 }\)

a_{1}= 0.6, a_{2} = 1.7

d = a_{2} – a_{1} = 1.7 – 0.6 = 1.1

So, first term a_{1} = 0.6 and common difference d = 1.1

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 + √2, 3 + 2√2, 3 + 3√2, …..

(vi) 0.2, 0.22, 0.222, 0.2222, ……

(vii) 0, -4, -8, -12, …..

(viii) \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , \(\frac { -1 }{ 2 }\) , …….

(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, 7², ……

Solution:

(i) 2, 4, 8, 16, ……

a_{2} – a_{1} = 4 – 2 = 2

a_{3} – a_{2} = 8 – 4 = 4

i.e., d = a_{n+1} – a_{n} is not same every time, so the given series do not form of an A.P.

(ii) We have given the series,

i.e., d = a_{n+1} – a_{n} an is same everything, so the given list of numbers form an AP.

So, common difference (d) = \(\frac { 1 }{ 2 }\)

Next three terms after the last given term are

\(\frac { 7 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = 4

4 + \(\frac { 1 }{ 2 }\) = \(\frac { 9 }{ 2 }\) and \(\frac { 9 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = 5

Thus, we obtain three terms as 4, \(\frac { 9 }{ 2 }\), 5.

(iii) We have given the series,

– 1.2, – 3.2, – 5.2, – 7.2, …

∴ a_{2} – a_{1} = – 3.2 – (1.2) = – 2

a_{3} – a_{2} = – 5.2 – (- 3.2) = – 2

i.e., d = a_{n+1} – a<sub<n is same every time, so the given list of numbers form an AP.

So, common difference (d) = – 2

Next three terms after the last term are,

– 7.2 + (- 2) = – 9.2

– 9.2 + (- 2) = – 11.2 and – 11.2 + (- 2), – 13.2

Thus, we obtain three terms as – 9.2, 11.2, – 13.2.

(iv) We have gives series,

10, – 6, -2, 2

∴ a_{2} – a_{1} = – 6 – (- 10) = 4

a_{3} – a_{2} = – 2 -( – 6) = 4

i.e., d = a_{n+1} is same every time, so the given list of numbers form an AP.

So, common difference (d) = 4

Next three terms after after last given term

are,

2 + 4= 6

6 + 4 = 10 and 10 + 4 = 14 Thus, we obtain three terms as 6,10,14.

(vi) We have given the series,

0.2, 0.22, 0.222, 0.2222,…

∴ a_{2} – a_{1} = 0.2 – 0.20 = 0.02

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

i.e., d = a_{n+1} is not same everytime, so the given series do not form an AP.

(vii) We have given the series,

0, – 4, – 8, – 12, …

= a_{2} – a_{1} = – 4 – 0 = – 4

a_{3} – a_{2} = – 8 – (- 4) = – 4

i. e., d = a_{n+1} – a_{n} is same everytime, so the given series is in the form of an AP.

So, common difference (d) = – 4

Next three terms after the last given terms

are,

– 12 + (- 4)= – 16

– 16 + (- 4) = – 20 and – 20 + (- 4) = – 24

Thus, we obtain three as – 16, – 20, – 24.

(viii) We have given the series,

i.e., d = a_{n+1} – a_{n} is same everytime, i.e., 0.

so the given series do not form an AP.

Next three terms after the last given terms are

\(\frac { -1 }{ 2 }\) + 0 = \(\frac { -1 }{ 2 }\), \(\frac { -1 }{ 2 }\) + 0 = \(\frac { -1 }{ 2 }\) and \(\frac { -1 }{ 2 }\) + 0 = \(\frac { -1 }{ 2 }\)

(ix) We have given the series, 1, 3, 9, 27,…

∴ a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 9 – 3 = 6

i.e., d = a_{n+1} – a_{n} is not always same.

So, the given series are, do not form an AP.

(x) We have given the series, a, 2a, 3a, 4a, …

∴ a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

i.e., d = a_{n+1} – a_{n} is same everytime, so the given series is in the form of an A.P.

So, common difference (d) = a

Next three terms after the last given term are, 4a + a = 5a

5a + a = 6a and 6a + a = 7a Thus, we obtain three terms as 5a, 6a, 7a.

(xi) We have given the series,

a_{1}, a², a³, a^{4}, ….

∴ a²_{2} – a_{1} = a (a – 1)

a³ – a² = a² (a – 1)

i.e., d = a_{n+1} – a_{n} is not always same.

So the given series are do not an A.P.

(xii) We have given the series,

i.e., d = a_{n+1} – a_{n} is same everytime so, the given series is in the form of an AP.

So, common difference (d) = \(\sqrt{2}\)

Next three terms after the last terms ate,

Thus, we obtain three terms as

\(\sqrt{50}\), \(\sqrt{72}\), \(\sqrt{98}\).

(xiii) We have gives the series

i.e., d = a_{n+1} – a_{n} is not same evervtime, so the given sereis are do not form an AP.

(xiv) We have given the series, 1², 3², 5², 7²

∴ a_{2} – a_{1} = 3² – 1² = 9 – 1 = 8

a_{3} – a_{2} = 5² – 32 = 25 – 9 = 16

i.e., d = a_{n+1} – a_{n} is not same everytime, so the given series do not form an AP.

(xv) We have given the series, 1², 5², 7², 7², …

∴ a_{2} – a_{1} = 5² – 1² = 25 – 1 = 24

a_{3} – a_{2} = 7² – 5² = 49 – 25 = 24

a_{4} – a_{3} = 7³ – 7² = 73 – 49 = 24

i.e., d = a_{n+1} – a_{n }is same everytime, so the given series are in the form of an AP.

So, common difference (d) = 24

Next three terms after the last term are,

73 + 24 = 97

97 + 24 = 121 and 121 + 24 = 145

Thus we obtain three terms as 97, 121, 145.