We know that,
For an A.P., a_n = a + (n − 1) d
a₁₇ = a + (17 − 1) d
a₁₇ = a + 16d
Similarly, a₁₀ = a + 9d
It is given that
a₁₇ − a₁₀ = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

Given A.P. is 3, 15, 27, 39, …
a = 3
d = a₂ − a₁ = 15 − 3 = 12
a₅₄ = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n^th term be 771.
a_n = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65^th term was 132 more than 54^th term.

Or

Let n^th term be 132 more than 54^th term.
n = 54 + 132/2
= 54 + 11 = 65^th term

Let the first term of these A.P.s be a₁ and a₂ respectively and the common difference of these A.P.s be d.
For first A.P.,
a₁₀₀ = a₁ + (100 − 1) d
= a₁ + 99d
a₁₀₀₀ = a₁ + (1000 − 1) d
a₁₀₀₀ = a₁ + 999d
For second A.P.,
a₁₀₀ = a₂ + (100 − 1) d
= a₂ + 99d
a₁₀₀₀ = a₂ + (1000 − 1) d
= a₂ + 999d
Given that, difference between
100^th term of these A.P.s = 100
Therefore, (a₁ + 99d) − (a₂ + 99d) = 100
a₁ − a₂ = 100 ... (i)
Difference between 1000^th terms of these A.P.s
(a₁ + 999d) − (a₂ + 999d) = a₁ − a₂
From equation (i),
This difference, a₁ − a₂ = 100
Hence, the difference between 1000^th terms of these A.P. will be 100.

First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
a_n = 994
n = ?
a_n = a + (n − 1) d
994 = 105 + (n − 1) 7
889 = (n − 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the n^th term of this A.P.
a = 12
d = 4
a_n = 248
a_n = a + (n - 1) d
248 = 12 + (n - 1) × 4
236/4 = n - 1
59  = n - 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.

Or

Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be n, a_n = 248
⇒ a + (n - 1) d = 248
⇒ 12 + (n - 1) × 4 = 248
⇒4(n - 1) = 248
⇒ n - 1 = 59
⇒ n = 60

63, 65, 67, …
a = 63
d = a₂ − a₁ = 65 − 63 = 2
n^th term of this A.P. = a_n = a + (n − 1) d
a_n= 63 + (n − 1) 2 = 63 + 2n − 2
a_n = 61 + 2n ... (i)
3, 10, 17, …
a = 3
d = a₂ − a₁ = 10 − 3 = 7
n^th term of this A.P. = 3 + (n − 1) 7
a_n = 3 + 7n − 7
a_n = 7n − 4 ... (ii)
It is given that, n^th term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13^th terms of both these A.P.s are equal to each other. 

a₃ = 16

a + (3 − 1) d = 16
a + 2d = 16 ... (i)
a₇ − a₅ = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …

Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a₂₀ = a + (20 − 1) d
a₂₀ = 253 + (19) (−5)
a₂₀ = 253 − 95
a = 158
Therefore, 20^th term from the last term is 158.