Class10 NCRT Arithmetic Progressions Exercise – 5.1 pdf || UP Board

 Arithmetic Progressions
Exercise – 5.1

 

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.

 

Fare for 1 km a₁ =  ₹15

Fare for 2 km a₂ =  ₹(15 + 8) = ₹23

Fare for 3 km a₃ =  ₹(23 + 8) = ₹31

Fare for 4 km a₄ =  ₹(31 + 8) = ₹39

a₂ – a₁ = 23 – 15 = 8

a₃ – a₂ = 31 – 23 = 8

a₄ – a₃ = 39 – 31 = 8

The difference of consecutive terms is the same. Hence, it is an A.P. ra.hmetic Progressions

 

(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

 

Let the amount of air present in the cylinder a₁ =  x

The remaining air in the cylinder after the first removal a₂ = x –  ¼ x = ¾ x

The remaining air in the cylinder after the second removal a₃ = ¾ x – ¼(¾ x) = 9/16 x

The remaining air in the cylinder after the third removal a₄ =

The difference of consecutive terms is not the same. Hence, it is not an A.P.

 

(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. 

 

1 meter digging cost a₁ =  ₹150

2 meter digging cost a₂ =  ₹(150 + 50) = ₹200

3 meter digging cost a₃ =  ₹(200 + 50) = ₹250

4 meter digging cost a₄ =  ₹(250 + 50) = ₹300

a₂ – a₁ = 200 – 150 = 50

a₃ – a₂ = 250 – 200 = 50

a₄ – a₃ = 300 – 250 = 50

The difference of consecutive terms is the same. Hence, it is an A.P.

 

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

 

Rate = 8% per annum, Principal a₁ = ₹10000

Amount after 1 year a₂ = ₹10000(1 + 8/100)¹ = ₹ 10800

Amount after 2 years a₃ = ₹10800(1 + 8/100)² = ₹ 10800(108/100)²

⇒ ₹12597

Amount after 3 years a₃ = ₹12597(1 + 8/100)³ = ₹10800(108/100)³

⇒ ₹15869

a₂ – a₁ = ₹10800 – ₹10000 = ₹800

a₃ – a₂ = ₹12597 – ₹10800 = ₹1797

a₄ – a₃ = ₹15869 – ₹12597 = ₹3272

The difference of consecutive terms is not the same. Hence, it is not an A.P.

 

2. Write first four terms of the A.P, When the first term a and the common difference d are given as follows:

(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 = ½

 

First term a₁ = a = -1

Second term a₂ = a₁ + d = -1 + ½  = - ½ 

Third term a₃ = a₂ + d = - ½  + ½ = 0

Fourth term a₄ = a₃ + d = 0 + ½ = ½

 

(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

 

3. For the following A.Ps, write the first term and the common difference:

 

(i) 3, 1, -1, -3,……

First term a = 3

Common difference d = a₂ - a₁ = 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

 

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,……

a₂ – a₁ = 4 – 2 = 2

a₃ – a₂ = 8 – 4 = 4

a₄ – a₃ = 16 – 8 = 8

The difference of consecutive terms is not the 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 of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  ½, next three terms of this AP are below-

Fifth term a₅ = a₄ + d = 7/2 + ½ = 4

Sixth term a₆ = a₅ + d = 4 + ½ = 9/2

Seventh term a₇ = a₆ + d = 9/2 + ½ = 10/2 = 5

 

(iii) -1.2, -3.2, -5.2, -7.2,…….

a₂ – a₁ = -3.2 – (-1.2) = -2.0

a₃ – a₂ = -5.2 – (-3.2) = -2.0

a₄ – a₃ = -7.2 – (-5.2) = -2.0

The difference of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  -2.0, next three terms of this AP are below-

Fifth term a₅ = a₄ + d = -7.2 – 2.0 = -9.2

Sixth term a₆ = a₅ + d = -9.2 – 2.0 = -11.2

Seventh term a₇ = a₆ + d = -11.2 – 2.0 = -13.2

 

(iv) -10, -6, -2, 2,…….

a₂ – a₁ = -6 – (-10) = 4

a₃ – a₂ = -2 – (-6) = 4

a₄ – a₃ = 2 – (-2) = 4

The difference of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  4, next three terms of this AP are below -

Fifth term a₅ = a₄ + d = 2 + 4 = 6

Sixth term a₆ = a₅ + d = 6 + 4 = 10

Seventh term a₇ = a₆ + d = 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 of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  √2, next three terms of this AP are below -

Fifth term a₅ = a₄ + d = (3 + 3√2) + √2 = 3 + 4√2

Sixth term a₆ = a₅ + d = (3 + 4√2) + √2  = 3 + 5√2

Seventh term a₇ = a₆ + d = (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 of consecutive terms is not the same. Hence, it is not an A.P.

 

(vii) 0, -4, -8, -12,…….

a₂ – a₁ = -4 – 0 = -4

a₃ – a₂ = -8 – (-4) = -8 + 4 = - 4

a₄ – a₃ = -12 – (-8) = -12 + 8 = - 4

The difference of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  - 4, next three terms of this AP are below -

Fifth term a₅ = a₄ + d = -12 - 4 = -16

Sixth term a₆ = a₅ + d = -16 - 4 = -20

Seventh term a₇ = a₆ + d = -20 - 4 = -24

 

The difference of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  0, next three terms of this AP are below -

 

(ix) 1, 3, 9, 27,……

a₂ – a₁ = 3 – 1 = 2

a₃ – a₂ = 9 – 3 = 6

a₄ – a₃ = 27 – 9 = 18

The difference of consecutive terms is not the 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 of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  a, next three terms of this AP are below -

Fifth term a₅ = a₄ + d = 4a + a = 5a

Sixth term a₆ = a₅ + d = 5a + a = 6a

Seventh term a₇ = a₆ + d = 6a + a = 7a

 

(xi) a, a², a³, a⁴,……..

a₂ – a₁ = a² – a

a₃ – a₂ = a³ – a²

a₄ – a₃ = a⁴ – a³ 

The difference of consecutive terms is not the 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 = √32 - √18 = 4√2 - 3√2 = √2

The difference of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  √2, next three terms of this AP are below -

Fifth term a₅ = a₄ + d = √32 + √2 = 4√2 + √2 = 5√2 = √(25×2) = √50

Sixth term a₆ = a₅ + d = 5√2 + √2  = 6√2 = √(36×2) = √72

Seventh term a₇ = a₆ + d = 6√2 + √2 = 7√2 = √(49×2) =√98

 

(xiii) √3, √6, √9, √12,…….

a₂ – a₁ = √6 – √3 

a₃ – a₂ = √9 – √6 = 3 - √6

a₄ – a₃ = √12 – √9 = 2√3 - 3

The difference of consecutive terms is not the same. Hence, it is not an A.P.

 

(xiv) 1², 3², 5², 7²,……….

a₂ – a₁ = 3² – 1² = 9 – 1 = 8

a₃ – a₂ = 5² – 3² = 25 – 9 = 16

a₄ – a₃ = 7² – 5² = 49 – 25 = 24

The difference of consecutive terms is not the same. Hence, it is not an A.P.

 

(xv) 1², 5², 7², 73,………

a₂ – a₁ = 5² – 1² = 25 – 1 = 24

a₃ – a₂ = 7² – 5² = 49 – 25 = 24

a₄ – a₃ = 73 – 7² = 73 – 49 = 24

The difference of consecutive terms is the same. Hence, it is an A.P.

Common difference d =  24, next three terms of this AP are below-

Fifth term a₅ = a₄ + d = 73 + 24 = 97

Sixth term a₆ = a₅ + d = 97 + 24 = 121

Seventh term a₇ = a₆ + d = 121 + 24 = 145

Download Class10 NCRT Arithmetic Progressions Exercise – 5.1 pdf