RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

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
a₁ = ₹ 15
a₂ = 15 + 8 = ₹23
a₃ = 23 + 8 = ₹31
Now, a₃ – a₂ = 31 – 23 = 8
a₂ – a₁ = 23 – 15 = 8
Here, a₃ – a₂ = a₂ – a₁ = 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, a₃ – a₂ ≠ a₂ – a₁
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,
a₁ = ₹150
a₂ = ₹ (150 + 50) = ₹ 200
a₃ = ₹ (200 + 50)
= ₹ 250 and so on so forth
Now, a₃ – a₂ = ₹ (250 – 200) = ₹ 50
a₂ – a₁ = ₹ (200 – 150) = ₹ 50
Here, a₃ – a₂ = a₂ – a₁ = 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
a₁ = ₹ 10,000
a₂ – ₹ 10,000 + \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,000 + 800 = 10,800
a₃ = ₹ 10,800 + \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,800 + 864
= ₹ 11,640 and so on so forth
Now a₃ – a₂ = ₹ (11,640 – 10,800) = ₹ 840
a₂ – a₁ = ₹ (10,800- 10,000) = ₹ 800
Here a₃ – a₂ ≠ a₂ – a₁
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
∴ a₁ = a = 10;
a₂ = a + d= 10 + 10 = 20;
a₃ = a + 2d = 10 + 2 × 10
= 10 + 20 = 30;
a₄ = 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
∴ a₁ = a = – 2
a₂= a + d = – 2 + 0 = – 2
a₃ = a +2d = -2 + 2 × 0 = -2
a₄ = 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
∴ a₁ = a = 4; a₂ = a + d = 4 – 3 = 1
a₃ = a + 2d = 4 + 2 (-3) = 4 – 6 = -2
a₄ = 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 = \(\frac{1}{2}\)
∴ a₁ = a = – 1; a₂ = a + d
= -1 + \(\frac{1}{2}\) = \(\frac{-1}{2}\)
a₃ = a + 2d = -1 + 2\(\left(\frac{1}{2}\right)\) = -1 + 1 = 0
a₄ = 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
∴ a₁ = a = -1.25
a₂ = a + d = – 1.25 – 0.25 = -1.50
a₃ = a + 2d = – 1.25 + 2 (- 0.25)
= – 1.25 – 0.50 = – 1.75
a₄ = 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, a₁ = 3, a₂ = 1, a₃ = – 1, a₄ = – 3,
first term = a₁ = 3
Now. a₂ – a₁ = 1 – 3 = -2
a₃ – a₂ = – 1 – 1 = – 2
a₄ – a₃ = – 3 + 1 = – 2
∴ a₂ – a₁ = a₃ – a₂ – a₄ – a₃ = – 2
Hence, the common difference = – 2 and
first term = 3

(ii) According to the question the AP is – 5, – 1, 3, 7, ….
Here, a₁ = – 5, a₂ = – 1, a₃ = 3, a₄ = 7
first term = a₁ = – 5
Now, a₂ – a₁ = -1 + 5 = 4
a₃ – a₂ = 3 + 1 = 4
a₃ – a₃ = 7 – 3 = 4
∴ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = 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 a₁ = 0.6, a₂ = 1.7, a₃ = 2.8, a₄ = 3.9
first term = a₁ = 0.6
Now, a₂ – a₁ = 1.7 – 0.6 = 1.1
a₃ – a₂ = 2.8 – 1.7 = 1.1
a₄ – a₃ = 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, a², a³, a⁴, ….
(xii) \(\sqrt {2}\), \(\sqrt {8}\), \(\sqrt {18}\), \(\sqrt {32}\), ….
(xiii) \(\sqrt {3\), \(\sqrt {6}\), \(\sqrt {9}\), \(\sqrt {12}\), ……..
(xiv) 1², 3², 5¹, 7², …..
(xv) 1², 5², 7², 7³, …..
Solution:
(i) The given temis are : 2, 4, 8, 16, ….
Here, a₁ = 2, a₂ = 4, a₃ = 8, a₄ = 16
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
∵ a₂ – a₁ ≠ a₃ – a₂
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, a₁ = – 1.2, a₂ = -3.2, a₃ = -5.2, a₄ = -7.2
a₂ – a₁ = -3.2 + 1.2 = -2
a₃ – a₂ = -5.2 + 3.2 = – 2
a₄ – a₃ = -7.2 + 5.2 = -2
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₂ =-2
Hence, the given list of numbers forms an AP.
∴ Common difference = d = – 2
Now, a₅ = a + 4d
= – 1.2 + 4 (-2)
= – 1.2 – 8 = – 9.2
a₆ = a + 5d = – 1.2 + 5 (-2)
= -1.2 – 10 = – 11.2
a₇ = 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, a₁ = – 10, a₂ = -6, a₃ = -2, a₄ = 2
a₂ – a₁ = -6 + 10 = 4
a₃ – a₂ = -2 + 6 = 4
a₄ – a₃ = 2 + 2 = 4
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₂ = 4
Hence, the given list of numbers forms an AP.
∴ Common difference = d = 4
Now, a₅ = a + 4d= – 10 + 4 (4)
= -10 + 16 = 6
a₆ = a + 5d = -10 + 5 (4)
= -10 + 20 = 10
a₇ = 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, a₁ = 3, a₂ = 3 + \(\sqrt {2}\),
a₃ =3 + 2\(\sqrt {2}\), a₄ = 3 + 3\(\sqrt {2}\)
a₂ – a₁ = 3 + \(\sqrt {2}\) – 3 = \(\sqrt {2}\)
a₃ – a₂ = 3 + 2\(\sqrt {2}\) – (3 + \(\sqrt {2}\))
= 3 + 2\(\sqrt {2}\) – 3 – \(\sqrt {2}\) = \(\sqrt {2}\)
a₄ – a₃ = 3 + 3\(\sqrt {2}\) – (3 + 2\(\sqrt {2}\))
= 3 + 3\(\sqrt {2}\) – 3 – 2\(\sqrt {2}\) = \(\sqrt {2}\)
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = \(\sqrt {2}\)
Hence, the given list of numbers form an AP.
∴ Common difference = d = \(\sqrt {2}\)
Nowr, a₅ = a + 4d = 3 + 4(\(\sqrt {2}\)) = 3 + 4\(\sqrt {2}\)
a₆ = a + 5d = 3 + 5\(\sqrt {2}\)
a₇ = 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, a₁ = 0.2, a₂ = 0.22,
a₃ = 0.222, a₄ = 0.2222
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ =0.222 – 0.22 = 0.002
a₄ – a₃ = 0.2222 – 0.222 = 0.0002
∵ a₂ – a₁ ≠ a₃ – a₂ ≠ a₄ – a₃
Hence, the given list of numbers do not form an AP.

(vii) The given terms are : 0, – 4, – 8, – 12, ….
Here, a₁ = 0, a₂ = – 4,
a₃ = -8, a₄ = – 12
a₂ – a₁ = -4 – 0 = -4
a₃ – a₂ = -8 + 4 = -4
a₄ – a₃ = – 12 + 8 = – 4
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
Hence, the given list of numbers form an AR
∴ Common difference = d = – 4
Now, a₅ = a + 4d = 0 + 4 (- 4) = – 16
a₆ = a + 5d = 0 + 5 (-4) = -20
a₇ = 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, a₁ = 1, a₂ = 3,
a₃ = 9, a₄ = 27
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
∵ a₂ – a₁ ≠ a₄ – a₂ ≠ a₄ – a₃
∴ The given list of numbers do not form an AP

(x) The given terms are : a, 2a, 3a, 4a, ….
Here, a₁ = a, a₂ = 2a,
a₃ = 3a, a₄ = 4a
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = a
Hence, the given list of numbers form an AP.
∴ Common difference = d = a
Now, a₅ = a + 4d=a + 4(a) = a + 4a = 5a
a₆ = a + 5d = a + 5a = 6a
a₇ = 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, a², a³, a⁴, ….
Here, a₁ = a, a₂ = a²,
a₃ = a³, a₄ = a⁴
a₂ – a₁ = a² – a = a(a – 1)
a₃ – a₂ = a³ – a² = a²(a – 1)
∵ a₂ – a₁ ≠ a₄ – a₂
∴ The given list of numbers do not form an AP.

(xii) The given terms are : \(\sqrt {2}\), \(\sqrt {8}\), \(\sqrt {18}\) , \(\sqrt {32}\), ….
Here, a₁ = \(\sqrt {2}\) , a₂ = \(\sqrt {8}\) , a₃ = \(\sqrt {18}\) a₄ = \(\sqrt {32}\)
or a₁ = \(\sqrt {2}\) , a₂ = 2\(\sqrt {2}\),
a₃ = 3\(\sqrt {2}\) , a₄ = 4\(\sqrt {2}\)
a₂ – a₁ = 2\(\sqrt {2}\) – \(\sqrt {2}\) = \(\sqrt {2}\)
a₃ – a₂ = 3\(\sqrt {2}\) – 2\(\sqrt {2}\) = \(\sqrt {2}\)
a₄ – a₃ = 4\(\sqrt {2}\) – 3\(\sqrt {2}\) = \(\sqrt {2}\)
V a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = \(\sqrt {2}\)
Hence, the given list of numbers form an AP.
Common difference = d= \(\sqrt {2}\)
Now, a₅ = a + 4d = \(\sqrt {2}\) + 4\(\sqrt {2}\) =5\(\sqrt {2}\)
a₆ = a + 5d = \(\sqrt {2}\) + 5\(\sqrt {2}\) = 6\(\sqrt {2}\)
a₇ = 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, a₁ = \(\sqrt {3}\), a₂ = \(\sqrt {6}\),
a₃ = \(\sqrt {9}\), a₄ = \(\sqrt {12}\)
or a₁ = \(\sqrt {3}\), a₂ = \(\sqrt {6}\),
a₃ = 3, a₄ = 2\(\sqrt {3}\)
a₂ – a₁ = \(\sqrt {6}\) – \(\sqrt {3}\) = \(\sqrt {3}\)(\(\sqrt {2}\) – 1)
Hence, a₃₂ – a₂ = \(\sqrt {9}\) – \(\sqrt {6}\) = \(\sqrt {3}\)(\(\sqrt {3}\) – \(\sqrt {2}\))
∴ The given list of numbers do not form an AP.
∵ a₂ – a₁ ≠ a₃ – a₂
Hence the given AP does not form an AP.

(xiv) The given terms are : 1², 3², 5², 7², …
Here, a₁ = a², a₂ = 3², a₃ = 5², a₄ = 7²
or a₁ = 1, a₂ = 9,
a₃ = 25, a₄ = 49
a₂ – a₁ = 9 – 1 = 8
a₃ – a₂ = 25 – 9 = 16
∵ a₂ – a₁ ≠ a₃ – a₂
∴ The given term do not form an AP.

(xv) The given terms are : l1², 5², 7², 73, ….
Here, a₁ = 1², a₂ = 5²,
a₃ = 7², a₄ = 73
or a₁ = 1, a₂ = 25,
a₃ = 49, a₄ = 73
a₂ – a₁ = 25 – 1 = 24
a₃ – a₂ = 49 – 25 = 24
a₄ – a₁ = 73 – 49 = 24
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = 24
Hence, the given list of numbers form an AP.
∴ Common difference = d = 24
Now, a₅ = a + 4d= 1 +4 (24) = 1 + 96 = 97
a₆ = a + 5d = 1 + 5 (24) = 1 + 120= 121
a₇ = a + 6d = 1 + 6(24) = 1 + 144= 145
Hence, the next three terms of the given AP are 97, 121 and 145.