Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions 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) Let the taxi fare of n^{th} km be given by a_{n}. Therefore according to die question

a_{1} = ₹ 15

a_{2} = 15 + 8 = ₹23

a_{3} = 23 + 8 = ₹31

Now, a_{3} – a_{2} = 31 – 23 = 8

a_{2} – a_{1} = 23 – 15 = 8

Here, a_{3} – a_{2} = a_{2} – a_{1} = 8

Therefore the given situation is of the form of AP.

(ii) Let the amount of air present in a cylinder be given by a_{n}. Therefore according to the question

Here, a_{3} – a_{2} ≠ a_{2} – a_{1}

The given situation is not of the form of AP.

(iii) Let the cost of digging n^{th} metre of a well be given by a_{n}. Therefore according to the question,

a_{1} = ₹150

a_{2} = ₹ (150 + 50) = ₹ 200

a_{3} = ₹ (200 + 50)

= ₹ 250 and so on so forth

Now, a_{3} – a_{2} = ₹ (250 – 200) = ₹ 50

a_{2} – a_{1} = ₹ (200 – 150) = ₹ 50

Here, a_{3} – a_{2} = a_{2} – a_{1} = 50

∴ The given situation is of die form of AP

(iv) Let the amount of n^{th} year be expressed by a_{n}. Therefore according to the question

a_{1} = ₹ 10,000

a_{2} – ₹ 10,000 + \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)

= ₹ 10,000 + 800 = 10,800

a_{3} = ₹ 10,800 + \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)

= ₹ 10,800 + 864

= ₹ 11,640 and so on so forth

Now a_{3} – a_{2} = ₹ (11,640 – 10,800) = ₹ 840

a_{2} – a_{1} = ₹ (10,800- 10,000) = ₹ 800

Here a_{3} – a_{2} ≠ a_{2} – a_{1}

The given situation is not of the form of AP.

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 = \(\frac{1}{2}\)

(v) a = – 1.25, d = – 0.25

Solution:

(i) According to the question

first term = a = 10

and common difference = d = 10

∴ a_{1} = a = 10;

a_{2} = a + d= 10 + 10 = 20;

a_{3} = a + 2d = 10 + 2 × 10

= 10 + 20 = 30;

a_{4} = a + 3d= 10 + 3 × 10

= 10 + 30 = 40

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

(ii) According to the question,

first term = a = – 2

and common difference = d = 0

∴ a_{1} = a = – 2

a_{2}= a + d = – 2 + 0 = – 2

a_{3} = a +2d = -2 + 2 × 0 = -2

a_{4} = a + 3d = -2 + 3 × 0 = -2

Hence, the first four terms of the AP are – 2, – 2, – 2, – 2

(iii) According to the question,

first term = a = 4

and common difference = d = -3

∴ a_{1} = a = 4; a_{2} = a + d = 4 – 3 = 1

a_{3} = a + 2d = 4 + 2 (-3) = 4 – 6 = -2

a_{4} = a + 3d = 4 + 3(-3) = 4 – 9 = -5

Hence, the first four terms of the AP are 4, 1, – 2, – 5

(iv) According to the question,

first term = a = – 1

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

∴ a_{1} = a = – 1; a_{2} = a + d

= -1 + \(\frac{1}{2}\) = \(\frac{-1}{2}\)

a_{3} = a + 2d = -1 + 2\(\left(\frac{1}{2}\right)\) = -1 + 1 = 0

a_{4} = a + 3d = -1 + 3\(\left(\frac{1}{2}\right)\)

= \(\frac{-2+3}{2}\) = \(\frac{1}{2}\)

Hence, the first four terms of the AP are – 1, \(\frac{-1}{2}\), 0, \(\frac{1}{2}\)

(v) According to the question,

first term = a = -1.25

and common difference -d = – 0.25

∴ a_{1} = a = -1.25

a_{2} = a + d = – 1.25 – 0.25 = -1.50

a_{3} = a + 2d = – 1.25 + 2 (- 0.25)

= – 1.25 – 0.50 = – 1.75

a_{4} = a + 3d = -1.25 + 3 (-0.25)

= – 1.25 -0.75 = -2

Hence, 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) According to the question, the AP is 3, 1, -1, -3, ….

Here, a_{1} = 3, a_{2} = 1, a_{3} = – 1, a_{4} = – 3,

first term = a_{1} = 3

Now. a_{2} – a_{1} = 1 – 3 = -2

a_{3} – a_{2} = – 1 – 1 = – 2

a_{4} – a_{3} = – 3 + 1 = – 2

∴ a_{2} – a_{1} = a_{3} – a_{2} – a_{4} – a_{3} = – 2

Hence, the common difference = – 2 and

first term = 3

(ii) According to the question the AP is – 5, – 1, 3, 7, ….

Here, a_{1} = – 5, a_{2} = – 1, a_{3} = 3, a_{4} = 7

first term = a_{1} = – 5

Now, a_{2} – a_{1} = -1 + 5 = 4

a_{3} – a_{2} = 3 + 1 = 4

a_{3} – a_{3} = 7 – 3 = 4

∴ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = 4

Hence, the common difference = 4

and first term = – 5

(iii) According to the question, the AP is \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\), ….

Hence, the common difference = \(\frac{4}{3}\)

and first term = \(\frac{1}{3}\)

(iv) According to the question, the AP is 0.6, 1.7, 2.8, 3.9, …..

Here a_{1} = 0.6, a_{2} = 1.7, a_{3} = 2.8, a_{4} = 3.9

first term = a_{1} = 0.6

Now, a_{2} – a_{1} = 1.7 – 0.6 = 1.1

a_{3} – a_{2} = 2.8 – 1.7 = 1.1

a_{4} – a_{3} = 3.9 – 2.8 = 1.1

Hence, the common difference =1.1

and first term = 0.6

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\(\sqrt {2}\) , ….

(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, a^{2}, a^{3}, a^{4}, ….

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

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

(xiv) 1^{2}, 3^{2}, 5^{1}, 7^{2}, …..

(xv) 1^{2}, 5^{2}, 7^{2}, 7^{3}, …..

Solution:

(i) The given temis are : 2, 4, 8, 16, ….

Here, a_{1} = 2, a_{2} = 4, a_{3} = 8, a_{4} = 16

a_{2} – a_{1} = 4 – 2 = 2

a_{3} – a_{2} = 8 – 4 = 4

∵ a_{2} – a_{1} ≠ a_{3} – a_{2}

Hence, the given list of numbers does not form an AP.

(ii) The given terms are : 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ……

Hence, the next three terms of the given AP are 4, \(\frac{9}{2}\) and 5.

(iii) The given terms are : – 1.2, – 3.2, – 5.2, -7.2, ……

Here, a_{1} = – 1.2, a_{2} = -3.2, a_{3} = -5.2, a_{4} = -7.2

a_{2} – a_{1} = -3.2 + 1.2 = -2

a_{3} – a_{2} = -5.2 + 3.2 = – 2

a_{4} – a_{3} = -7.2 + 5.2 = -2

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{2} =-2

Hence, the given list of numbers forms an AP.

∴ Common difference = d = – 2

Now, a_{5} = a + 4d

= – 1.2 + 4 (-2)

= – 1.2 – 8 = – 9.2

a_{6} = a + 5d = – 1.2 + 5 (-2)

= -1.2 – 10 = – 11.2

a_{7} = a + 6d = – 1.2 + 6 (- 2)

= – 1.2 – 12 = – 13.2

Hence, the next three terms of the given AP are – 9.2, – 11.2 and – 13.2

(iv) The given terms are : – 10, -6, -2, 2, …..

Here, a_{1} = – 10, a_{2} = -6, a_{3} = -2, a_{4} = 2

a_{2} – a_{1} = -6 + 10 = 4

a_{3} – a_{2} = -2 + 6 = 4

a_{4} – a_{3} = 2 + 2 = 4

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{2} = 4

Hence, the given list of numbers forms an AP.

∴ Common difference = d = 4

Now, a_{5} = a + 4d= – 10 + 4 (4)

= -10 + 16 = 6

a_{6} = a + 5d = -10 + 5 (4)

= -10 + 20 = 10

a_{7} = a + 6d = -10 + 6 (4)

= -10 + 24 = 14

Hence, the next three terms of the given AP are 6, 10 and 14.

(v) The given terms are : 3,3 + \(\sqrt {2}\),3 + 2\(\sqrt {2}\) ,3 + 3\(\sqrt {2}\) , ….

Here, a_{1} = 3, a_{2} = 3 + \(\sqrt {2}\),

a_{3} =3 + 2\(\sqrt {2}\), a_{4} = 3 + 3\(\sqrt {2}\)

a_{2} – a_{1} = 3 + \(\sqrt {2}\) – 3 = \(\sqrt {2}\)

a_{3} – a_{2} = 3 + 2\(\sqrt {2}\) – (3 + \(\sqrt {2}\))

= 3 + 2\(\sqrt {2}\) – 3 – \(\sqrt {2}\) = \(\sqrt {2}\)

a_{4} – a_{3} = 3 + 3\(\sqrt {2}\) – (3 + 2\(\sqrt {2}\))

= 3 + 3\(\sqrt {2}\) – 3 – 2\(\sqrt {2}\) = \(\sqrt {2}\)

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = \(\sqrt {2}\)

Hence, the given list of numbers form an AP.

∴ Common difference = d = \(\sqrt {2}\)

Nowr, a_{5} = a + 4d = 3 + 4(\(\sqrt {2}\)) = 3 + 4\(\sqrt {2}\)

a_{6} = a + 5d = 3 + 5\(\sqrt {2}\)

a_{7} = a + 6d = 3 + 6\(\sqrt {2}\)

Hence, the next three terms of the given AP are 3 + 4\(\sqrt {2}\), 3 + 5\(\sqrt {2}\) and 3 + 6\(\sqrt {2}\)

(vi) The given terms are : 0.2, 0.22, 0.222, 0.2222, ……

Here, a_{1} = 0.2, a_{2} = 0.22,

a_{3} = 0.222, a_{4} = 0.2222

a_{2} – a_{1} = 0.22 – 0.2 = 0.02

a_{3} – a_{2} =0.222 – 0.22 = 0.002

a_{4} – a_{3} = 0.2222 – 0.222 = 0.0002

∵ a_{2} – a_{1} ≠ a_{3} – a_{2} ≠ a_{4} – a_{3}

Hence, the given list of numbers do not form an AP.

(vii) The given terms are : 0, – 4, – 8, – 12, ….

Here, a_{1} = 0, a_{2} = – 4,

a_{3} = -8, a_{4} = – 12

a_{2} – a_{1} = -4 – 0 = -4

a_{3} – a_{2} = -8 + 4 = -4

a_{4} – a_{3} = – 12 + 8 = – 4

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3}

Hence, the given list of numbers form an AR

∴ Common difference = d = – 4

Now, a_{5} = a + 4d = 0 + 4 (- 4) = – 16

a_{6} = a + 5d = 0 + 5 (-4) = -20

a_{7} = a + 6d = 0 + 6 (- 4) = – 24

Hence, the next three terms of the given AP are – 16,-20 and – 24.

(viii) The given terms are :

Hence, the next three terms of the given sequence are –\(\frac{1}{2}\), –\(\frac{1}{2}\) and –\(\frac{1}{2}\)

(ix) The given terms are : 1, 3, 9, 27, ….

Here, a_{1} = 1, a_{2} = 3,

a_{3} = 9, a_{4} = 27

a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 9 – 3 = 6

a_{4} – a_{3} = 27 – 9 = 18

∵ a_{2} – a_{1} ≠ a_{4} – a_{2} ≠ a_{4} – a_{3}

∴ The given list of numbers do not form an AP

(x) The given terms are : a, 2a, 3a, 4a, ….

Here, a_{1} = a, a_{2} = 2a,

a_{3} = 3a, a_{4} = 4a

a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

a_{4} – a_{3} = 4a – 3a = a

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = a

Hence, the given list of numbers form an AP.

∴ Common difference = d = a

Now, a_{5} = a + 4d=a + 4(a) = a + 4a = 5a

a_{6} = a + 5d = a + 5a = 6a

a_{7} = a + 6d = a + 6a = 7a

Hence, the next three terms of the given AP are 5a. 6a and 7a.

(xi) The given terms are : a, a^{2}, a^{3}, a^{4}, ….

Here, a_{1} = a, a_{2} = a^{2},

a_{3} = a^{3}, a_{4} = a^{4}

a_{2} – a_{1} = a^{2} – a = a(a – 1)

a_{3} – a_{2} = a^{3} – a^{2} = a^{2}(a – 1)

∵ a_{2} – a_{1} ≠ a_{4} – a_{2}

∴ The given list of numbers do not form an AP.

(xii) The given terms are : \(\sqrt {2}\), \(\sqrt {8}\), \(\sqrt {18}\) , \(\sqrt {32}\), ….

Here, a_{1} = \(\sqrt {2}\) , a_{2} = \(\sqrt {8}\) , a_{3} = \(\sqrt {18}\) a_{4} = \(\sqrt {32}\)

or a_{1} = \(\sqrt {2}\) , a_{2} = 2\(\sqrt {2}\),

a_{3} = 3\(\sqrt {2}\) , a_{4} = 4\(\sqrt {2}\)

a_{2} – a_{1} = 2\(\sqrt {2}\) – \(\sqrt {2}\) = \(\sqrt {2}\)

a_{3} – a_{2} = 3\(\sqrt {2}\) – 2\(\sqrt {2}\) = \(\sqrt {2}\)

a_{4} – a_{3} = 4\(\sqrt {2}\) – 3\(\sqrt {2}\) = \(\sqrt {2}\)

V a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = \(\sqrt {2}\)

Hence, the given list of numbers form an AP.

Common difference = d= \(\sqrt {2}\)

Now, a_{5} = a + 4d = \(\sqrt {2}\) + 4\(\sqrt {2}\) =5\(\sqrt {2}\)

a_{6} = a + 5d = \(\sqrt {2}\) + 5\(\sqrt {2}\) = 6\(\sqrt {2}\)

a_{7} = a + 6d= \(\sqrt {2}\) + 6\(\sqrt {2}\) = 7\(\sqrt {2}\)

Hence, the next three terms of the given AP are 5\(\sqrt {2}\), 6\(\sqrt {2}\) and 7\(\sqrt {2}\).

(xiii) The given terms are : \(\sqrt {3}\), \(\sqrt {6}\), \(\sqrt {9}\), \(\sqrt {12}\), …

Here, a_{1} = \(\sqrt {3}\), a_{2} = \(\sqrt {6}\),

a_{3} = \(\sqrt {9}\), a_{4} = \(\sqrt {12}\)

or a_{1} = \(\sqrt {3}\), a_{2} = \(\sqrt {6}\),

a_{3} = 3, a_{4} = 2\(\sqrt {3}\)

a_{2} – a_{1} = \(\sqrt {6}\) – \(\sqrt {3}\) = \(\sqrt {3}\)(\(\sqrt {2}\) – 1)

Hence, a_{32} – a_{2} = \(\sqrt {9}\) – \(\sqrt {6}\) = \(\sqrt {3}\)(\(\sqrt {3}\) – \(\sqrt {2}\))

∴ The given list of numbers do not form an AP.

∵ a_{2} – a_{1} ≠ a_{3} – a_{2}

Hence the given AP does not form an AP.

(xiv) The given terms are : 1^{2}, 3^{2}, 5^{2}, 7^{2}, …

Here, a_{1} = a^{2}, a_{2} = 3^{2}, a_{3} = 5^{2}, a_{4} = 7^{2}

or a_{1} = 1, a_{2} = 9,

a_{3} = 25, a_{4} = 49

a_{2} – a_{1} = 9 – 1 = 8

a_{3} – a_{2} = 25 – 9 = 16

∵ a_{2} – a_{1} ≠ a_{3} – a_{2}

∴ The given term do not form an AP.

(xv) The given terms are : l1^{2}, 5^{2}, 7^{2}, 73, ….

Here, a_{1} = 1^{2}, a_{2} = 5^{2},

a_{3} = 7^{2}, a_{4} = 73

or a_{1} = 1, a_{2} = 25,

a_{3} = 49, a_{4} = 73

a_{2} – a_{1} = 25 – 1 = 24

a_{3} – a_{2} = 49 – 25 = 24

a_{4} – a_{1} = 73 – 49 = 24

∵ a_{2} – a_{1} = a_{3} – a_{2} = a_{4} – a_{3} = 24

Hence, the given list of numbers form an AP.

∴ Common difference = d = 24

Now, a_{5} = a + 4d= 1 +4 (24) = 1 + 96 = 97

a_{6} = a + 5d = 1 + 5 (24) = 1 + 120= 121

a_{7} = a + 6d = 1 + 6(24) = 1 + 144= 145

Hence, the next three terms of the given AP are 97, 121 and 145.

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