Ans. (i)Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23 

Taxi fare after 3 km = 23 + 8 = Rs 31

Taxi fare after 4 km = 31 + 8 = Rs 39

Therefore, the sequence is 15, 23, 31, 39…

It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …)

(ii)Let amount of air initially present in a cylinder = V

Amount of air left after pumping out air by vacuum pump = 

Amount of air left when vacuum pump again pumps out air

= 

So, the sequence we get is like 

Checking for difference between consecutive terms …

Difference between consecutive terms is not equal.

Therefore, it is not an arithmetic progression.

(iii) Cost of digging 1 meter of well = Rs 150

Cost of digging 2 meters of well = 150 + 50 = Rs 200

Cost of digging 3 meters of well = 200 + 50 = Rs 250

Therefore, we get a sequence of the form 150, 200, 250…

It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50…)

Here, difference between any two consecutive terms which is also called common difference is equal to 50.

(iv)Amount in bank after Ist year = … (1)

Amount in bank after two years = … (2)

Amount in bank after three years = … (3)

Amount in bank after four years = … (4)

It is not an arithmetic progression because (2) − (1) ≠ (3) − (2)

(Difference between consecutive terms is not equal)

Therefore, it is not an Arithmetic Progression.

Ans. (i) First term = a = 10, d = 10 

Second term = a + d = 10 + 10 = 20

Third term = second term + d = 20 + 10 = 30

Fourth term = third term + d = 30 + 10 = 40

Therefore, first four terms are: 10, 20, 30, 40

(ii) First term = a = –2 , d = 0

Second term = a + d = –2 + 0 = –2

Third term = second term + d = –2 + 0 = –2

Fourth term = third term + d = –2 + 0 = –2

Therefore, first four terms are: –2, –2, –2, –2

(iii) First term = a = 4, d =–3

Second term = a + d = 4 – 3 = 1

Third term = second term + d = 1 – 3 = –2

Fourth term = third term + d = –2 – 3 = –5

Therefore, first four terms are: 4, 1, –2, –5

(iv) First term = a = –1, d = 

Second term = a + d = –1 +  = −

Third term = second term + d = −+ = 0

Fourth term = third term + d = 0 + =

Therefore, first four terms are: –1, −, 0,

(v) First term = a = –1.25, d = –0.25

Second term = a + d = –1.25 – 0.25 = –1.50

Third term = second term + d = –1.50 – 0.25 = –1.75

Fourth term = third term + d

= –1.75 – 0.25 = –2.00

Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00

Ans. (i) 3, 1, –1, –3… 

First term = a = 3,

Common difference (d) = Second term – first term = Third term – second term and so on

Therefore, Common difference (d) = 1 – 3 = –2

(ii) –5, –1, 3, 7…

First term = a = –5

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = –1 – (–5) = –1 + 5 = 4

(iii)

First term = a = 

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = 

(iv) 0.6, 1.7, 2.8, 3.9…

First term = a = 0.6

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = 1.7 − 0.6 = 1.1

Ans. (i) 2, 4, 8, 16… 

It is not an AP because difference between consecutive terms is not equal.

As4 – 2 ≠ 8 − 4

(ii)2, , 3, …

It is an AP because difference between consecutive terms is equal.

Common difference (d) = 

Fifth term =  Sixth term = 4 + ½ = 

Seventh term = 

Therefore, next three terms are 4, and 5.

(iii)−1.2, −3.2, −5.2, −7.2…

It is an AP because difference between consecutive terms is equal.

 −3.2 − (−1.2)

= −5.2 − (−3.2)

= −7.2 − (−5.2) = −2

Common difference (d) = −2

Fifth term = −7.2 – 2 = −9.2Sixth term = −9.2 – 2 = −11.2

Seventh term = −11.2 – 2 = −13.2

Therefore, next three terms are −9.2, −11.2 and −13.2

(iv) −10, −6, −2, 2…

It isan AP because difference between consecutive terms is equal.

 −6 − (−10) = −2 − (−6)

= 2 − (−2) = 4

Common difference (d) = 4

Fifth term = 2 + 4 = 6 Sixth term = 6 + 4 = 10

Seventh term = 10 + 4 = 14

Therefore, next three terms are 6, 10 and 14

(v) 

It is an AP because difference between consecutive terms is equal.

 

Common difference (d) = 

Fifth term = 

Sixth term = 

Seventh term = 

Therefore, next three terms are 

(vi) 0.2, 0.22, 0.222, 0.2222…

It is not an AP because difference between consecutive terms is not equal.

0.22 − 0.2 ≠ 0.222 − 0.22

(vii) 0, −4, −8, −12…

It is an AP because difference between consecutive terms is equal.

 −4 – 0 = −8 − (−4)

= −12 − (−8) = −4

Common difference (d) = −4

Fifth term = −12 – 4 =−16 Sixth term = −16 – 4 = −20

Seventh term = −20 – 4 = −24

Therefore, next three terms are −16, −20 and −24

(viii) 

It is an AP because difference between consecutive terms is equal.

 

Common difference (d) = 0

Fifth term = Sixth term = 

Seventh term = 

Therefore, next three terms are 

(ix) 1, 3, 9, 27…

It is not an AP because difference between consecutive terms is not equal.

3 – 1 ≠ 9 − 3

(x) a, 2a, 3a, 4a…

It is an AP because difference between consecutive terms is equal.

 2a – a = 3a − 2a = 4a − 3a = a

Common difference (d) = a

Fifth term = 4a + a = 5a Sixth term = 5a + a = 6a

Seventh term = 6a + a = 7a

Therefore, next three terms are 5a, 6a and 7a

(xi) a, a², a³, a⁴…

It is not an AP because difference between consecutive terms is not equal.

a² – a ≠ a³ − a²

(xii) 

 

It is an AP because difference between consecutive terms is equal.

 

Common difference (d) = 

Fifth term =  Sixth term = 

Seventh term = 

Therefore, next three terms are 

(xiii) 

It is not an AP because difference between consecutive terms is not equal.

(xiv) 

It is not an AP because difference between consecutive terms is not equal.

(xv) 

 1, 25, 49, 73…

It is an AP because difference between consecutive terms is equal.

 

= = 24

Common difference (d) = 24

Fifth term = 73 + 24 = 97 Sixth term = 97 + 24 = 121

Seventh term = 121 + 24 = 145

Therefore, next three terms are 97, 121 and 145