(i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a₂ − a₁ = 7 − 10 = −3
We know that, a_n = a + (n − 1) d
a₃₀ = 10 + (30 − 1) (−3)
a₃₀ = 10 + (29) (−3)
a₃₀ = 10 − 87 = −77
Hence, the correct answer is option C.

(ii) Given that A.P. is -3, -1/2, ,2 ...
First term a = - 3
Common difference, d = a₂ − a₁ = (-1/2) - (-3)
= (-1/2) + 3 = 5/2
We know that, a_n = a + (n − 1) d
a₁₁ = 3 + (11 -1)(5/2)
a₁₁ = 3 + (10)(5/2)
a₁₁ = -3 + 25
a₁₁ = 22
Hence, the answer is option B.

(i) For this A.P.,

a = 2
a₃ = 26
We know that, a_n = a + (n − 1) d
a₃ = 2 + (3 - 1) d
26 = 2 + 2d
24 = 2d
d = 12
a₂ = 2 + (2 - 1) 12
= 14
Therefore, 14 is the missing term.

(ii) For this A.P.,
a₂ = 13 and
a₄ = 3
We know that, a_n = a + (n − 1) d
a₂ = a + (2 - 1) d
13 = a + d ... (i)
a₄ = a + (4 - 1) d
3 = a + 3d ... (ii)
On subtracting (i) from (ii), we get
- 10 = 2d
d = - 5
From equation (i), we get
13 = a + (-5)
a = 18
a₃ = 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,
a = 5 and
a₄ = 19/2
We know that, a_n = a + (n − 1) d
a₄ = a + (4 - 1) d
19/2 = 5 + 3d
19/2 - 5 = 3d3d = 9/2
d = 3/2

a₂ = a + (2 - 1) d
a₂ = 5 + 3/2
a₂ = 13/2

a₃ = a + (3 - 1) d

a₃ = 5 + 2×3/2

a₃ = 8

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a = −4 and
a₆ = 6
We know that,
a_n = a + (n − 1) d
a₆ = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a₂ = a + d = − 4 + 2 = −2
a₃ = a + 2d = − 4 + 2 (2) = 0
a₄ = a + 3d = − 4 + 3 (2) = 2
a₅ = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v)
For this A.P.,
a₂ = 38
a₆ = −22
We know that
a_n = a + (n − 1) d
a₂ = a + (2 − 1) d
38 = a + d ... (i)
a₆ = a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d 
d = −15
a = a₂ − d = 38 − (−15) = 53
a₃ = a + 2d = 53 + 2 (−15) = 23
a₄ = a + 3d = 53 + 3 (−15) = 8
a₅ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

3, 8, 13, 18, …
For this A.P.,
a = 3
d = a₂ − a₁ = 8 − 3 = 5
Let n^th term of this A.P. be 78.
a_n = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16^th term of this A.P. is 78.

For this A.P.,
a = 11
d = a₂ − a₁ = 8 − 11 = −3
Let −150 be the n^th term of this A.P.
We know that,
a_n = a + (n − 1) d
-150 = 11 + (n - 1)(-3)
-150 = 11 - 3n + 3
-164 = -3n
n = 164/3
Clearly, n is not an integer.
Therefore, - 150 is not a term of this A.P.

Given that,
a₁₁ = 38
a₁₆ = 73
We know that,
a_n = a + (n − 1) d
a₁₁ = a + (11 − 1) d
38 = a + 10d ... (i)
Similarly,
a₁₆ = a + (16 − 1) d
73 = a + 15d ... (ii)
On subtracting (i) from (ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a₃₁ = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^st term is 178.

Given that,
a₃ = 12
a₅₀ = 106
We know that,
a_n = a + (n − 1) d
a₃ = a + (3 − 1) d
12 = a + 2d ... (i)
Similarly, a₅₀ = a + (50 − 1) d
106 = a + 49d ... (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a₂₉ = a + (29 − 1) d
a₂₉ = 8 + (28)2
a₂₉ = 8 + 56 = 64
Therefore, 29^th term is 64.

Given that,

a₃ = 4
a₉ = −8
We know that,
a_n = a + (n − 1) d
a₃ = a + (3 − 1) d
4 = a + 2d ... (i)
a₉ = a + (9 − 1) d
−8 = a + 8d ... (ii)
On subtracting equation (i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = a + 2 (−2)
4 = a − 4
a = 8
Let n^th term of this A.P. be zero.
a_n = a + (n − 1) d
0 = 8 + (n − 1) (−2)
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, 5^th term of this A.P. is 0.