# Gujarat Board Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Gujarat Board Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

## 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 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.

(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

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

a_{1} = a

a_{1} = 10

a_{2} = a + d

a_{2} = 10 + 10 = 20

a_{3} = a + 2d = 10 + 2 x 10

= 10 + 20 = 30

a_{4} = a + 3d = 10 + 3 x 10

= 10 + 30 = 40

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

(ii) As a_{1} = 4 and d = – 3

a_{2} = a + d = 4 + (- 3) = 1

a_{3} = a + 2d = 4 + 2 x ( – 3)

= 4 – 6 = – 2

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

= 4 – 9 = – 5

So the first four terms of the AP are: 4, 1, -2, -5.

(iii) As a_{1} = a = – 2 and d = 0

a_{2} = a + d = – 2 + 0 = -2

a_{3} = a + 2d = – 2 + 2 x 0 = – 2

a_{4} = a + 3d = – 2 + 3 x 0 = – 2

Therefore the first four terms of the AP are:

– 2, – 2, – 2, – 2.

(v) As a = – 1.25 and d = – 0.25

a_{1} = a = – 1.25

a_{2} = a + d = – 1.25 + (-0.25)

= – 1.25 – 0.25 = – 1.50

a_{3} = a + 2d = – 1.25 + 2 x (- 0.25)

= – 1.25 – 0.50

a_{3} = – 1.75

a_{4} = 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) 1/3, 5/3, 9/3, 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 = a_{2} – a_{1}

= 1 – 3 = – 2

d = – 2

(ii) As the given AP is – 5, – 1, 3, 7, ………….

Here first term a = – 5

and Common difference d = a_{2} – a_{1}

= – 1 – (- 5)

d = – 1 + 5 = 4

Common difference is 4.

(iv) As the given AP is 0.6, 1.7, 2.8, 3.9, ……………

Here a = 0.6

and Common difference d = a_{2} – a_{1}

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.

(iii) The given sequence is -1.2, -3.2, -5.2, -7.2,

a_{3} – a_{1} = – 3.2 – (- 1.2)

= – 3.2 + 1.2 = – 2

a_{3} – a_{2} = – 5.2 – (- 3.2)

= – 5.2 + 3.2 = – 2

a_{4} – a_{3} = – 7.2 – (- 5.2)

= – 7.2 + 5.2 – 2

Common difference = – 2 is same so the

given sequence is in AP.

a_{5}= a + 4d = -1.2 + 4( – 2)

= – 1.2 – 8 = – 9.2

a_{6} = 0 +5d = – 1.2 + 5 x (- 2)

= – 1.2 – 10 = – 11.2

and a_{7} = 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, …………..

a_{2} – a_{1} = – 6 – ( – 10) = – 6 + 10 = 4

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

a_{4} – a_{3} = 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.

a_{5} = a + 4d = -10 + 4 x 4

= – 10 + 16 = 6

a_{6} = a + 5d = – 10 + 5 x 4

= – 10 + 20 = 10

a_{7} = a + 6d = – 10 + 6 x 4

= – 10 + 24 = 14

(vi) Given sequence is 0.2, 0.22, 0.222, 0.2222, ……….

a_{2} – a_{1} = 0.22 – 0.2 = 0.02

a_{3} – a_{2} = 0.222 – 0.22 = 0.002

Since a_{2} – a_{1} ≠ a_{3} – a_{2} therefore given sequence does not form an AP.

(vii) Given sequence is 0, – 4, – 8, – 12

a_{2} – a_{1} = – 4 – 0 = – 4

a_{3} – a_{2} = – 8 – ( – 4) = – 4

a_{4} – a_{3}= – 12 – ( – 8) = – 4

Since difference between two consecutive

terms is same therefore the sequence is in AP.

Next three terms are

(ix) Given sequence is 1, 3, 9, 27, ………

a_{2} – a_{1} = 3 – 1 = 2

a_{3} – a_{2} = 9 – 3 = 6

Since a_{2} – a_{1} ≠ a_{3} – a_{2} therefore given sequence does not form an AP.

(x) Given sequence is a, 2a, 3a, 4a, …..

a_{2} – a_{1} = 2a – a = a

a_{3} – a_{2} = 3a – 2a = a

a_{4} – a_{3} = 4a – 3a = a

Since common difference is same. Hence the given sequence is in AP.

a_{1} = a

d = a

Next three terms of AP are

a_{5} = a + 4d = a + 4a = 5a

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

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

(xi) Given sequence is a, a^{2}, a^{3}, a^{4}, …………..

a_{2} – a_{1} = a^{2} – = a (a – 1)

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

Since a_{2} – a_{1} ≠ a_{3} – a_{2}therefore given sequence is not in AP.

(xiv) Given sequence is 1^{2}, 5^{2}, 7^{2}, …………

a_{2} – a_{1} = 3^{2} – 1^{2} = 9 – 1 = 8

a_{3} – a_{2} = 5^{2} – 3^{2} = 9 – 1 = 16

Since a_{2} – a_{1} ≠ a_{3} – a_{2} therefore the given sequence does not form an AP.

(xv) Given sequence is 1^{2}, 5^{2}, 73, …………

a = 1^{2} = 1

a_{2} – a_{1} = 5^{2} – 1^{2} = 25 – 1 = 24

a_{3} – a_{2} = 7^{2} – 5^{2} = 49 – 25 = 24

a_{4} – a_{3} = 73 – 7^{2} = 73 – 49 = 24

Since common difference is same then sequence is in AP.

Next terms are

a_{5} = a + 4d = 1 + 4 x 24 = 97

a_{6} = a + 5d 1 + 5 x 24 = 121

a_{7} = a + 6d = 1 + 6 x 24 = 145.

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