GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Gujarat Board GSEB Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths 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 kilometre.
(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 cost ₹150, for the first metre and rises by ₹5O 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) We can observed that
taxi fare for first km = ₹15
taxi fare for first 2 km = ₹15 + ₹8 = ₹23
taxi fare for first 3 km = 15 + 2 x 8 = 15 + 16 = ₹31
taxi fare for first 4 km = 15 + 3 x 8
= 15 + 24 = 39
Hence the terms are 15, 23,31, 39 ……………. these terms form an AP because every term is 8 more than the preceding terms.

(ii) Let us suppose the initial volume of air in a cylinder by \(\frac { 1 }{ 4 } \)υ litre.
In each stroke vacuum pump removes
then volume of air in cylinder after
First stroke = υ – \(\frac { 1 }{ 4 } \) – υ = \(\frac { 3υ }{ 4 } \)
Volume of air remaining in cylinder after second stroke
= \(\frac { 3υ }{ 4 } \) – \(\frac { 1 }{ 4 } \)\(\left( \frac { 3v }{ 4 } \right) \)
= \(\frac { 3υ }{ 4 } \) – \(\frac { 3υ }{ 16 } \) = \(\frac { 12υ-3υ }{ 16 } \)
= \(\frac { 9υ }{ 16 } \)
Volume of air remaining in cylinder after
Third stroke = \(\frac { 9υ }{ 16 } \) – \(\frac { 1 }{ 4 } \) x \(\frac { 9υ }{ 16 } \)
= \(\frac { 9υ }{ 16 } \) \(\left[ 1-\frac { 1 }{ 4 } \right] \)
\(\frac { 9υ }{ 16 } \) x \(\frac { 3 }{ 4 } \) = \(\frac { 27 }{ 64 } \)υ
Volume of air remaining in cylinder after
Fourth stroke = \(\frac { 27υ }{ 64 } \) – \(\frac { 1 }{ 4 } \)\(\left( \frac { 27υ }{ 64 } \right) \)
= \(\frac { 27 }{ 64 } \)υ\(\left[ 1-\frac { 1 }{ 4 } \right] \)
= \(\frac { 27 }{ 64 } \) x \(\frac { 3 }{ 4 } \) = \(\frac { 81υ }{ 256 } \)
thus the terms are υ, \(\frac { 3υ }{ 4 } \), \(\frac { 9υ }{ 16 } \), \(\frac { 27 }{ 64 } \)υ, \(\frac { 81 }{ 256 } \)υ
Note that a₂ – a₁ = \(\frac { 3υ }{ 4 } \)υ – \(\frac { -υ }{ 4 } \)
Also, a₃ – a₂ = \(\frac { 9υ }{ 16 } \) – \(\frac { 3υ }{ 4 } \) = \(\frac { 9υ-12υ }{ 16 } \)
= \(\frac { -3υ }{ 16 } \)
a₂ – a₁ ≠ a₃ – a₂
Hence the difference between two consecutive terms are not same.
So the above sequence does not form an AP.

(iii) It is given that
the cost of digging for first metre = 150
the cost of digging for first 2 metres = a + d
= 150 + 50 = 200
the cost of digging for first 3 metres
= a + 2d = 250
= 150 + 2 x 50
the cost of digging for first 4 metres = (a + 3d)
= 150 + 3 x 50 = 300
We observe that, the term obtained 150, 200, 250. 300 form an AP because every term is 50 more than the preceding term.

(iv) Let P is the amount be deposted at r% compound interest per annum for n years then after n years money becomes
= P\(\left( 1+\frac { r }{ 100 } \right) \)ⁿ
Now, here P = 10000 and r = 8%
So amount after 1 years= 10000 \(\left( 1+\frac { 8 }{ 100 } \right) \)¹
Amount after 2 years= 10000\(\left( 1+\frac { 8 }{ 100 } \right) \)²
Amount after 3 years = 10000\(\left( 1+\frac { 8 }{ 100 } \right) \)³
We have the sequence 10000\(\left( 1+\frac { 8 }{ 100 } \right) \)¹, \(\left( 1+\frac { 8 }{ 100 } \right) \)², \(\left( 1+\frac { 8 }{ 100 } \right) \)³, \(\left( 1+\frac { 8 }{ 100 } \right) \)⁴, ………….
We observe that the difference between two consecutive terms is not same. So the list of numbers is not in 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 = 4, d = – 3
(iii) a = – 2, d = 0
(iv) a = -2, d = \(\frac { 1 }{ 2 } \)
(v) a = – 1.25, d = – 0.25
Solution:
(i) It is given that a = 10, d = 10
First four terms of the AP are a₁, a₂, a₃, a₄,
a₁ = a
a₁ = 10
a₂ = a + d
a₂ = 10 + 10 = 20
a₃ = a + 2d = 10 + 2 x 10
= 10 + 20 = 30
a₄ = a + 3d = 10 + 3 x 10
= 10 + 30 = 40
Therefore first four terms of the AP are: 10, 20, 30, 40.

(ii) As a₁ = 4 and d = – 3
a₂ = a + d = 4 + (- 3) = 1
a₃ = a + 2d = 4 + 2 x ( – 3)
= 4 – 6 = – 2
a₄ = a + 3d = 4+ 3 x (-3)
= 4 – 9 = – 5
So the first four terms of the AP are: 4, 1, -2, -5.

(iii) As a₁ = a = – 2 and d = 0
a₂ = a + d = – 2 + 0 = -2
a₃ = a + 2d = – 2 + 2 x 0 = – 2
a₄ = a + 3d = – 2 + 3 x 0 = – 2
Therefore the first four terms of the AP are:
– 2, – 2, – 2, – 2.

(iv) As n – 1 and
then a₁ = a = – 1
a₂ = a + d = – 1 + \(\frac { 1 }{ 2 } \)
\(\frac { -2+1 }{ 2 } \) = \(\frac { -1 }{ 2 } \)
a₃ = a + 2d = – 1 + 2 x \(\frac { 1 }{ 2 } \)
= – 1 + 1 = 0
a₄ = a + 3d = – 1 + 3 x \(\frac { 1 }{ 2 } \)
\(\frac { -2+3 }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Hence the first four terms of the AP are : – 1, \(\frac { -1 }{ 2 } \), 0, \(\frac { 1 }{ 2 } \).

(v) As a = – 1.25 and d = – 0.25
a₁ = a = – 1.25
a₂ = a + d = – 1.25 + (-0.25)
= – 1.25 – 0.25 = – 1.50
a₃ = a + 2d = – 1.25 + 2 x (- 0.25)
= – 1.25 – 0.50
a₃ = – 1.75
a₄ = a + 3d = – 1.25 + 3 x (- 0.25)
= – 1.25 – 0.75 = – 2.00
Hence the first four terms of the AP are:
– 1.25, – 1.50, – 1.75, – 2.00.

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) As the given AP is 3, 1,-1,-3, …..
Here first terms a = 3
and Common difference d = a₂ – a₁
= 1 – 3 = – 2
d = – 2

(ii) As the given AP is – 5, – 1, 3, 7, ………….
Here first term a = – 5
and Common difference d = a₂ – a₁
= – 1 – (- 5)
d = – 1 + 5 = 4
Common difference is 4.

(iii) Asthegiven AP is \(\frac { 1 }{ 3 } \), \(\frac { 5 }{ 3 } \), \(\frac { 9 }{ 3 } \), \(\frac { 13 }{ 3 } \)
Here first term a = \(\frac { 1 }{ 3 } \)
3
and Common difference d = a₂ – a₁
= \(\frac { 5 }{ 3 } \) – \(\frac { 1 }{ 3 } \) = \(\frac { 4 }{ 3 } \)
= \(\frac { 4 }{ 3 } \)
Common difference is \(\frac { 4 }{ 3 } \)

(iv) As the given AP is 0.6, 1.7, 2.8, 3.9, ……………
Here a = 0.6
and Common difference d = a₂ – a₁
d = 1.7 – 0.6 = 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+\(\sqrt { 2 } \), 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, 5², 7², …………
(xv) 1², 5², 7², 7², ………….
Solution:
(i) The given list of numbers is: 2, 4, 8, 16
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
Here a₂ – a₁ ≠ a₃ – a₂
Hence the given sequence is not in AP.

(ii) Here the given list of numbers is 2, \(\frac { 5 }{ 2 } \), 3, \(\frac { 7 }{ 2 } \), …………..
a₂ – a₁ = \(\frac { 5 }{ 2 } \) – 2 = \(\frac { 1 }{ 2 } \)
a₃ – a₂ = 3 – \(\frac { 5 }{ 2 } \) = \(\frac { 1 }{ 2 } \)
a₄ – a₃ = \(\frac { 7 }{ 2 } \) – 3 = \(\frac { 1 }{ 2 } \)
Hence common difference, d = \(\frac { 1 }{ 2 } \) the common difference of given sequence is same so the given sequence is in AP.
The next three terms are
a₅ = a + 4\(\frac { 1 }{ 2 } \) = 2 + 4 = 2 + 2 = 4
a₆ = a + 5d = 2 + 5 x \(\frac { 1 }{ 2 } \)
= 4+5 9
a₇ = a + 6d = 2 + 6 x \(\frac { 1 }{ 2 } \)
= 2 + 3 = 5
Hence next three terms of the AP are 4, \(\frac { 9 }{ 2 } \), 5

(iii) The given sequence is -1.2, -3.2, -5.2, -7.2,
a₃ – a₁ = – 3.2 – (- 1.2)
= – 3.2 + 1.2 = – 2
a₃ – a₂ = – 5.2 – (- 3.2)
= – 5.2 + 3.2 = – 2
a₄ – a₃ = – 7.2 – (- 5.2)
= – 7.2 + 5.2 – 2
Common difference = – 2 is same so the
given sequence is in AP.
a₅= a + 4d = -1.2 + 4( – 2)
= – 1.2 – 8 = – 9.2
a₆ = 0 +5d = – 1.2 + 5 x (- 2)
= – 1.2 – 10 = – 11.2
and a₇ = a + 6d = – 1.2 + 6 x (- 2)
= – 1.2 – 12 = – 13.2
Hence, three terms of AP are – 9.2, – 11.2, -13.2

(iv) Given sequence is – 10, – 6, – 2, 2, …………..
a₂ – a₁ = – 6 – ( – 10) = – 6 + 10 = 4
a₃ – a₂ = – 2 – ( – 6) = – 2 + 6 = 4
a₄ – a₃ = 2 – ( – 2) = 2 + 2 = 4
Since given sequence has same common
difference. Hence the given sequence is in AP.
Next three terms of AP are.
a₅ = a + 4d = -10 + 4 x 4
= – 10 + 16 = 6
a₆ = a + 5d = – 10 + 5 x 4
= – 10 + 20 = 10
a₇ = a + 6d = – 10 + 6 x 4
= – 10 + 24 = 14

(v) Given sequence is 3, 3 + \(\sqrt { 2 } \), 3 + 2\(\sqrt { 2 } \), 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 – a₁ = 3 +3\(\sqrt { 2 } \) – (3 + 2\(\sqrt { 2 } \))
= 3 + 3\(\sqrt { 2 } \) – 3 – 2\(\sqrt { 2 } \) = \(\sqrt { 2 } \)
The difference between each two consecutive terms is same.
Hence the given sequence is an AP.
Next three terms are
a₅ = a + 4d = 3 + 4\(\sqrt { 2 } \)
a₆ = a + Sd = 3 + 5\(\sqrt { 2 } \)
a₇= a + 6d = 3 + 6\(\sqrt { 2 } \)

(vi) Given sequence is 0.2, 0.22, 0.222, 0.2222, ……….
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
Since a₂ – a₁ ≠ a₃ – a₂ therefore given sequence does not form an AP.

(vii) Given sequence is 0, – 4, – 8, – 12
a₂ – a₁ = – 4 – 0 = – 4
a₃ – a₂ = – 8 – ( – 4) = – 4
a₄ – a₃= – 12 – ( – 8) = – 4
Since difference between two consecutive
terms is same therefore the sequence is in AP.
Next three terms are
a₅=a + 4d = 0 + 4 x ( – 4) = – 16
a₆= a + 5d = 0 + 5 x ( – 4) = – 20
a₇=a + 6d = 0 + 6 x ( – 4) = – 24
Given sequence is \(\frac { -1 }{ 2 } \), \(\frac { -1 }{ 2 } \), \(\frac { -1 }{ 2 } \), \(\frac { -1 }{ 2 } \), ………………
a₂ – a₁ = \(\frac { -1 }{ 2 } \) – \(\left( \frac { -1 }{ 2 } \right) \) = \(\frac { -1 }{ 2 } \) + \(\frac { 1 }{ 2 } \) = 0
a₃ – a₂ = \(\frac { -1 }{ 2 } \) – \(\left( \frac { -1 }{ 2 } \right) \) = \(\frac { -1 }{ 2 } \) + \(\frac { 1 }{ 2 } \) = 0
a₄ – a₃ = \(\frac { -1 }{ 2 } \) – \(\left( \frac { -1 }{ 2 } \right) \) = \(\frac { -1 }{ 2 } \) + \(\frac { 1 }{ 2 } \) = 0
Since difference of each two consecutive terms is same therefore the sequence is in AP.
Next three terms are
a₅ = a + 4d = \(\frac { -1 }{ 2 } \) + 4 x 0 = \(\frac { -1 }{ 2 } \)
a₆ = a + 5d = \(\frac { -1 }{ 2 } \) + 5 x 0 = \(\frac { -1 }{ 2 } \)
a₇ = a + 6d = \(\frac { -1 }{ 2 } \) + 6 x 0 = \(\frac { -1 }{ 2 } \)

(ix) Given sequence is 1, 3, 9, 27, ………
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
Since a₂ – a₁ ≠ a₃ – a₂ therefore given sequence does not form an AP.

(x) Given sequence is a, 2a, 3a, 4a, …..
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
Since common difference is same. Hence the given sequence is in AP.
a₁ = a
d = a
Next three terms of AP are
a₅ = a + 4d = a + 4a = 5a
a₆ = a + 5d = a + 5a = 6a
a₇ = a + 6d = a + 6a = 7a

(xi) Given sequence is a, a², a³, a⁴, …………..
a₂ – a₁ = a² – = a (a – 1)
a₃ – a₂ = a³ – a² = a² (a – 1)
Since a₂ – a₁ ≠ a₃ – a₂therefore given sequence is not in AP.

(xii) Given sequence is \(\sqrt { 2 } \), \(\sqrt { 8 } \), \(\sqrt { 18 } \), \(\sqrt { 32 } \), …………..
Here a = \(\sqrt { 2 } \)
a₂ – ₁ = \(\sqrt { 8 } \) – \(\sqrt { 2 } \) = 2\(\sqrt { 2 } \) – \(\sqrt { 2 } \)
a₂ – a₁ = \(\sqrt { 2 } \)
a₃ – a₂ = \(\sqrt { 18 } \) – \(\sqrt { 8 } \) = 3\(\sqrt { 2 } \) – 2\(\sqrt { 2 } \)
a₃ – a₂ = \(\sqrt { 2 } \)
a₄ – a₃ = \(\sqrt { 32 } \) – \(\sqrt { 18 } \) = 4\(\sqrt { 2 } \) – 3\(\sqrt { 2 } \)
a₄ – a₃ = \(\sqrt { 2 } \)
Since common difference is same therefore sequence is in AP.
Hence next three terms are
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\(\sqrt { 2 } \)
5\(\sqrt { 2 } \), 6\(\sqrt { 2 } \), 7\(\sqrt { 2 } \)
i.e \(\sqrt { 50 } \), \(\sqrt { 72 } \), \(\sqrt { 98 } \)

(xiii) Given sequence is \(\sqrt { 3 } \), \(\sqrt { 6 } \), \(\sqrt { 9 } \), \(\sqrt { 12 } \), …………….
a₂ – a₁ = \(\sqrt { 6 } \) – \(\sqrt { 3 } \) = \(\sqrt { 3 } \) x \(\sqrt { 2 } \) – \(\sqrt { 3 } \)
= \(\sqrt { 3 } \) (\(\sqrt { 2 } \) – 1)
a₃ – a₂ = \(\sqrt { 9 } \) – \(\sqrt { 6 } \)
= \(\sqrt { 3 } \) x \(\sqrt { 3 } \) – \(\sqrt { 3 } \) x \(\sqrt { 2 } \)
= \(\sqrt { 3 } \) (\(\sqrt { 3 } \) – \(\sqrt { 2 } \))
Since a₂ – a₁ ≠ a₃ – a₂ therefore the given sequence does not form an AP.

(xiv) Given sequence is 1², 5², 7², …………
a₂ – a₁ = 3² – 1² = 9 – 1 = 8
a₃ – a₂ = 5² – 3² = 9 – 1 = 16
Since a₂ – a₁ ≠ a₃ – a₂ therefore the given sequence does not form an AP.

(xv) Given sequence is 1², 5², 73, …………
a = 1² = 1
a₂ – a₁ = 5² – 1² = 25 – 1 = 24
a₃ – a₂ = 7² – 5² = 49 – 25 = 24
a₄ – a₃ = 73 – 7² = 73 – 49 = 24
Since common difference is same then sequence is in AP.
Next terms are
a₅ = a + 4d = 1 + 4 x 24 = 97
a₆ = a + 5d 1 + 5 x 24 = 121
a₇ = a + 6d = 1 + 6 x 24 = 145.