# Arithmetic ProgressionsExercise – 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