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NCERT solutions for class 10 Maths chapter Arithmetic Progressions (Exercise 5.1)

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NCERT solutions for class 10 Maths 


Chapter 5


Arithmetic Progressions


Exercise 5.1


Ex 5.1 Question 1.


 In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

(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 costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution:

(i) Taxi fare for 1 km a1 = ₹ 15

Taxi fare for first 2 km a2 = ₹ 15+8 = 23

Taxi fare for first 3 km a3 = ₹ 23+8 = 31

Taxi fare for first 4 km a4 = ₹ 31+8 = 39

a2 – a1 = 23 – 15 = 8

a3 – a2 = 31 – 23 = 8

a4 – a3 = 39 – 31 = 8

The difference between the terms is same i.e. 8. Hence, it is an A.P.

(ii) Let the volume of air in a cylinder a1 = x

Amount of air left , after first removal a= x – 1/4x = 3/4x

Amount of air left , after second removal a3 = 3/4x – 1/4(3/4x) = 9/16x

Amount of air left , after third removal a4 = 9/16x – 1/4(9/16x) = 27/64x

a2 – a1 = 3/4x – x = -1/4x

a3 – a2 = 9/16x- 3/4x = -3/16x

a4 – a3 = 27/64x – 9/16x = -9/64x

The difference between the terms are not same.Hence, it is not an A.P.

(iii) Cost of digging 1metre deep a1 = ₹ 150

Cost of digging  2 metres deep a2= ₹ 150+50 = Rs.200

Cost of digging 3 metres deep a3 = ₹ 200+50 = Rs.250

Cost of digging  4 metres a4 =₹ 250+50 = Rs.300

a2 – a1 = 200 – 150 = 50

a3 – a2 = 250- 200 = 50

a4 – a3 = 300 – 250 = 50

The difference between the terms is same i.e. 50. Hence, it is an A.P.

(iv) Rate = 8% per annum, Principal a1 = ₹ 10,000

Amount after 1 year a2 = ₹ 10,000 (1+8/100)¹ = ₹ 10,800

Amount after 2 year a3 = ₹ 10,000 (1+8/100)² = ₹ 12,597

Amount after 3 year a4 = ₹ 10,000 (1+8/100)³ = ₹ 15,869

a2 – a1 = ₹10,800 – ₹10,000 = ₹800

a3 – a2 = ₹ 12,597 – ₹ 10,800 = ₹1797

a4 – a3 = ₹ 15,869 – ₹ 12,597 = ₹3292

The difference between the terms are not same.Hence, it is not an A.P.


Ex 5.1 Question 2.


Write first four terms of the A.P. when the first term a and the common difference are given as follows:

(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25

Solutions:

(i) a = 10, d = 10

First term a1 = a = 10

Second term a2 = a1+d = 10+10 = 20

Third term a3 = a2+d = 20+10 = 30

Fourth term a4 = a3+d = 30+10 = 40

(ii) a = – 2, d = 0

First term  a1 = a = -2

Second term  a2 = a1+d = – 2+0 = – 2

Third term  a3 = a2+d = – 2+0 = – 2

Fourth term a4 = a3+d = – 2+0 = – 2

(iii) a = 4, d = – 3

First term a1 = a = 4

Second term  a2 = a1+d = 4-3 = 1

Third term a3 = a2+d = 1-3 = – 2

Fourth term  a4 = a3+d = -2-3 = – 5

(iv) a = – 1, d = 1/2

First term a1 = a = -1

Second term a2 = a1+d = -1+1/2 = -1/2

Third term a3 = a2+d = -1/2+1/2 = 0

Fourth term a4 = a3+d = 0+1/2 = 1/2

(v) a = – 1.25, d = – 0.25

First term a1 = a = – 1.25

Second term a2 = a1 + d = – 1.25-0.25 = – 1.50

Third term a3 = a2 + d = – 1.50-0.25 = – 1.75

Fourth term a4 = a3 + d = – 1.75-0.25 = – 2.00


Ex 5.1 Question 3.


For the following A.P.s, 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) 3, 1, – 1, – 3 …

First-term, a = 3

Common difference, d = a2 – a= 1 – 3 = -2

(ii)  – 5, – 1, 3, 7 …

First term, a = -5

Common difference, d = a2 – a1 = ( – 1)-( – 5) = – 1+5 = 4

(iii) 1/3, 5/3, 9/3, 13/3 ….

First term, a = 1/3

Common difference, d = a2 – a1 = 5/3 – 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 …

First term, a = 0.6

Common difference, d = a2 – a1 =1.7 – 0.6 =1.1


Ex 5.1 Question 4.


Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.

(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 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) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) aa2a3a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …

Solution:

(i) 2, 4, 8, 16 …

a2 – a1 = 4 – 2 = 2

a3 – a2 = 8 – 4 = 4

a4 – a3 = 16 – 8 = 8

The difference between the terms are not same.Hence, it is not an A.P.

(ii) 2, 5/2, 3, 7/2 ….

a2 – a1 = 5/2-2 = ½

a3 – a2 = 3-5/2 = ½

a4 – a3 = 7/2-3 = ½

The difference between the terms is same i.e. ½. Hence, it is an A.P.

The next three terms –

a5 = 7/2+1/2 = 4

a6 = 4 +1/2 = 9/2

a7 = 9/2 +1/2 = 5

(iii) -1.2, – 3.2, -5.2, -7.2 …

a2 – a1 = (-3.2)-(-1.2) = -2

a3 – a2 = (-5.2)-(-3.2) = -2

a4 – a3 = (-7.2)-(-5.2) = -2

The difference between the terms is same i.e. -2. Hence, it is an A.P.

The next three terms –

a5 = – 7.2-2 = -9.2

a6 = – 9.2-2 = – 11.2

a7 = – 11.2-2 = – 13.2

(iv)-10, – 6, – 2, 2 …

a2 – a1 = (-6)-(-10) = 4

a3 – a2 = (-2)-(-6) = 4

a4 – a3 = (2 -(-2) = 4

The difference between the terms is same i.e. 4. Hence, it is an A.P.

The next three terms-

a5 = 2+4 = 6

a6 = 6+4 = 10

a7 = 10+4 = 14

(v) 3, 3+√2, 3+2√2, 3+3√2

a2 – a1 = 3+√2-3 = √2

a3 – a2 = (3+2√2)-(3+√2) = √2

a4 – a3 = (3+3√2) – (3+2√2) = √2

The difference between the terms is same i.e. √2. Hence, it is an A.P.

The next three terms –

a5 = (3+√2) +√2 = 3+4√2

a6 = (3+4√2)+√2 = 3+5√2

a7 = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 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

The difference between the terms are not same.Hence, it is not an A.P.

(vii) 0, -4, -8, -12 …
a2 – a1 = (-4)-0 = -4

a3 – a2 = (-8)-(-4) = -4

a4 – a3 = (-12)-(-8) = -4

The difference between the terms is same i.e. -4. Hence, it is an A.P.

The next three terms –

a5 = -12-4 = -16

a6 = -16-4 = -20

a7 = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

a2 – a1 = (-1/2) – (-1/2) = 0

a3 – a2 = (-1/2) – (-1/2) = 0

a4 – a3 = (-1/2) – (-1/2) = 0

The difference between the terms is same i.e. 0. Hence, it is an A.P.

The next three terms –

a5 = (-1/2)-0 = -1/2

a6 = (-1/2)-0 = -1/2

a7 = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

a2 – a1 = 3-1 = 2

a3 – a2 = 9-3 = 6

a4 – a3 = 27-9 = 18

The difference between the terms are not same.Hence, it is not an A.P.

(x) a, 2a, 3a, 4a …

a2 – a1 = 2a–a = a

a3 – a2 = 3a-2a = a

a4 – a3 = 4a-3a = a

The difference between the terms is same i.e. a. Hence, it is an A.P.

The next three terms –

a5 = 4a+a = 5a

a6 = 5a+a = 6a

a7 = 6a+a = 7a

(xi) aa2a3a4 …

a2 – a1 = a2–a = a(a-1)

a3 – a2 = a a= a2(a-1)

a4 – a3 = a4 – a= a3(a-1)

The difference between the terms are not same.Hence, it is not an A.P.

(xii) √2, √8, √18, √32 …


a2 – a1 = √8-√2  = 2√2-√2 = √2

a3 – a2 = √18-√8 = 3√2-2√2 = √2

a4 – a3 = 4√2-3√2 = √2

The difference between the terms is same i.e. √2. Hence, it is an A.P.

The next three terms –

a5 = √32+√2 = 4√2+√2 = 5√2 = √50

a6 = 5√2+√2 = 6√2 = √72

a7 = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)

a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)

a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

The difference between the terms are not same.Hence, it is not an A.P.

(xiv) 12, 32, 52, 72 …

or 1, 9, 25, 49 …..

a2 − a1 = 9−1 = 8

a3 − a= 25−9 = 16

a4 − a3 = 49−25 = 24

The difference between the terms are not same.Hence, it is not an A.P.

(xv) 12, 52, 72, 73 …

or 1, 25, 49, 73 …

a2 − a1 = 25−1 = 24

a3 − a= 49−25 = 24

a4 − a3 = 73−49 = 24

The difference between the terms is same i.e. 24. Hence, it is an A.P.

The next three terms –

a5 = 73+24 = 97

a6 = 97+24 = 121

a= 121+24 = 145


 

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