It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 - 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)² , (3V/4)³...
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

We know that if Rs P is deposited at r % compound interest per annum for n years, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(i) a = 10, d = 10
Let the series be a₁, a₂, a₃, a₄, a₅ …
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20
a₃ = a₂ + d = 20 + 10 = 30
a₄ = a₃ + d = 30 + 10 = 40
a₅ = a₄ + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = - 2, d = 0
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = -2
a₂ = a₁ + d = - 2 + 0 = - 2
a₃ = a₂ + d = - 2 + 0 = - 2
a₄ = a₃ + d = - 2 + 0 = - 2
Therefore, the series will be - 2, - 2, - 2, - 2 …
First four terms of this A.P. will be - 2, - 2, - 2 and - 2.

(iii) a = 4, d = - 3
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = 4
a₂ = a₁ + d = 4 - 3 = 1
a₃ = a₂ + d = 1 - 3 = - 2
a₄ = a₃ + d = - 2 - 3 = - 5
Therefore, the series will be 4, 1, - 2 - 5 …
First four terms of this A.P. will be 4, 1, - 2 and - 5.

(iv) a = - 1, d = 1/2
Let the series be a₁, a₂, a₃, a₄ …a₁ = a = -1
a₂ = a₁ + d = -1 + 1/2 = -1/2
a₃ = a₂ + d = -1/2 + 1/2 = 0
a₄ = a₃ + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = - 1.25, d = - 0.25
Let the series be a₁, a₂, a₃, a₄ …
a₁ = a = - 1.25
a₂ = a₁ + d = - 1.25 - 0.25 = - 1.50
a₃ = a₂ + d = - 1.50 - 0.25 = - 1.75
a₄ = a₃ + d = - 1.75 - 0.25 = - 2.00
Clearly, the series will be 1.25, - 1.50, - 1.75, - 2.00 ……..
First four terms of this A.P. will be - 1.25, - 1.50, - 1.75 and - 2.00.

(i) 3, 1, - 1, - 3 …
Here, first term, a = 3
Common difference, d = Second term - First term
= 1 - 3 = - 2

(ii) - 5, - 1, 3, 7 …
Here, first term, a = - 5
Common difference, d = Second term - First term
= ( - 1) - ( - 5) = - 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ....
Here, first term, a = 1/3

Common difference, d = Second term - First term 

= 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term - First term
= 1.7 - 0.6
= 1.1

(i) 2, 4, 8, 16 …
Here,
a₂ - a₁ = 4 - 2 = 2
a₃ - a₂ = 8 - 4 = 4
a₄ - a₃ = 16 - 8 = 8
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ....
Here,

a₂ - a₁ = 5/2 - 2 = 1/2
a₃ - a₂ = 3 - 5/2 = 1/2
a₄ - a₃ = 7/2 - 3 = 1/2
⇒ a_n+1 - a_n is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
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 …
Here,
a₂ - a₁ = ( -3.2) - ( -1.2) = -2
a₃ - a₂ = ( -5.2) - ( -3.2) = -2
a₄ - a₃ = ( -7.2) - ( -5.2) = -2
⇒ a_n+1 - a_n is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a₅ = - 7.2 - 2 = - 9.2
a₆ = - 9.2 - 2 = - 11.2
a₇ = - 11.2 - 2 = - 13.2

(iv) -10, - 6, - 2, 2 …
Here,
a₂ - a₁ = (-6) - (-10) = 4
a₃ - a₂ = (-2) - (-6) = 4
a₄ - a₃ = (2) - (-2) = 4
⇒ a_n+1 - a_n is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a₅ = 2 + 4 = 6
a₆ = 6 + 4 = 10
a₇ = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a₂ - a₁ = 3 + √2 - 3 = √2
a₃ - a₂ = (3 + 2√2) - (3 + √2) = √2
a₄ - a₃ = (3 + 3√2) - (3 + 2√2) = √2
⇒ a_n+1 - a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
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 ….
Here,
a₂ - a₁ = 0.22 - 0.2 = 0.02
a₃ - a₂ = 0.222 - 0.22 = 0.002
a₄ - a₃ = 0.2222 - 0.222 = 0.0002
⇒ a_n+1 - a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …
Here,
a₂ - a₁ = (-4) - 0 = -4
a₃ - a₂ = (-8) - (-4) = -4
a₄ - a₃ = (-12) - (-8) = -4
⇒ a_n+1 - a_n is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a₅ = -12 - 4 = -16
a₆ = -16 - 4 = -20
a₇ = -20 - 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ....
Here,
a₂ - a₁ = (-1/2) - (-1/2) = 0
a₃ - a₂ = (-1/2) - (-1/2) = 0
a₄ - a₃ = (-1/2) - (-1/2) = 0
⇒ a_n+1 - a_n is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a₅ = (-1/2) - 0 = -1/2
a₆ = (-1/2) - 0 = -1/2
a₇ = (-1/2) - 0 = -1/2

(ix) 1, 3, 9, 27 …
Here,
a₂ - a₁ = 3 - 1 = 2
a₃ - a₂ = 9 - 3 = 6
a₄ - a₃ = 27 - 9 = 18
⇒ a_n+1 - a_n is not the same every time.
Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …
Here,
a₂ - a₁ = 2a - a = a
a₃ - a₂ = 3a - 2a = a
a₄ - a₃ = 4a - 3a = a
⇒ a_n+1 - a_n is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a₅ = 4a + a = 5a
a₆ = 5a + a = 6a
a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴ …
Here,
a₂ - a₁ = a² - a = (a - 1)
a₃ - a₂ = a³ - a² = a² (a - 1)
a₄ - a₃ = a⁴ - a³ = a³(a - 1)
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 ...
Here,
a₂ - a₁ = √8 - √2  = 2√2 - √2 = √2
a₃ - a₂ = √18 - √8 = 3√2 - 2√2 = √2
a₄ - a₃ = 4√2 - 3√2 = √2
⇒ a_n+1 - a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
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 ...
Here,
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)
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(xiv) 1², 3², 5², 7² …

Or, 1, 9, 25, 49 …..
Here,
a₂ − a₁ = 9 − 1 = 8
a₃ − a₂ = 25 − 9 = 16
a₄ − a₃ = 49 − 25 = 24
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(xv) 1², 5², 7², 73 …
Or 1, 25, 49, 73 …
Here,
a₂ − a₁ = 25 − 1 = 24
a₃ − a₂ = 49 − 25 = 24
a₄ − a₃ = 73 − 49 = 24
i.e., a_k+1 − a_k is same every time.
⇒ a_n+1 - a_n is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a₅ = 73+ 24 = 97
a₆ = 97 + 24 = 121
a₇ = 121 + 24 = 145

(i) a = 7, d = 3, n = 8, a_n = ?
We know that,
For an A.P. a_n = a + (n − 1) d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, a_n = 28

(ii) Given that
a = −18, n = 10, a_n = 0, d = ?
We know that,
a_n = a + (n − 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2

(iii) Given that
d = −3, n = 18, a_n = −5
We know that,
a_n = a + (n − 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51
a = 51 − 5 = 46
Hence, a = 46

(iv) a = −18.9, d = 2.5, a_n = 3.6, n = ?
We know that,
a_n = a + (n − 1) d
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5
(n - 1) = 22.5/2.5
n - 1 = 9
n = 10
Hence, n = 10

(v) a = 3.5, d = 0, n = 105, a_n = ?
We know that,
a_n = a + (n − 1) d
a_n = 3.5 + (105 − 1) 0
a_n = 3.5 + 104 × 0
a_n = 3.5
Hence, a_n = 3.5